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+{
+ "chapter-atomic-structure": [
+ {
+ "title": "Chem Sec 1",
+ "originalName": "Section A - Single Correct",
+ "questions": [
+ {
+ "question_id": "atomic-structure-chem-sec-1-1-1",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": null,
+ "subject": "chemistry",
+ "chapter": "atomic-structure",
+ "chapterTitle": "Atomic Structure",
+ "type": "mcq",
+ "rawChapterType": "MCQ",
+ "originalNumber": 1,
+ "displayNumber": 1,
+ "image": "./assets/chapter-packs/atomic-structure/Chemistry Section 1__--__1__--__1.png",
+ "solutionImage": null,
+ "question": {
+ "content": "![\"Atomic](\"./assets/chapter-packs/atomic-structure/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: e m ratio of cathode rays is independent to the nature of gas.
Answer: B
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Solution: e m e m e e m mA A B B \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 A B \u21d2 2 3 3 2 = \u00d7 e e A B \u21d2 e e A B = 4 9
Answer: B
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Solution: eV = 1 2 2 m v \u21d2 v e m = \u00d7 \u00d7 2 V = \u00d7 \u00d7 \u00d7 = \u00d7 2 1 764 10 200 8 2 10 11 6 . . m/s
Answer: C
\nOriginal PDF solution page
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Solution: e m e m \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = mesor -particle \u03b1 1 1 1836 208 2 4 17 65 1 .
Answer: D
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Solution: eV = = 1 2 2 2 2 mv p m \u21d2 p m = 2 eV \u2234 p p p e = = 1836 1 42 85 1 .
Answer: A
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Solution: \u03c5 \u03c5 = = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 c s z 3 10 2 10 50 1 2 10 10 6 1 15 cm cm H .
Answer: A
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Solution: p E t n hc t = = \u22c5 \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u03bb 6 10 6 626 10 3 10 1 662 6 10 15 34 8 9 . . = 1.8 \u00d7 10 \u20133 J/s \u2013 m 2
Answer: D
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Solution: p E t n hc t = = \u22c5 \u22c5 \u03bb \u21d2 14 100 200 6 626 10 3 10 1 1987 8 7 10 34 8 9 \u00d7 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 n . . \u2234 n = 4 \u00d7 10 19 s \u20131
Answer: D
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Solution: E n hc hc = \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u03bb ( . N ) A 1 75 10 2500 10 84 4 10 J
Answer: B
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Solution: E x E abs \u00d7 = 100 emit \u21d2 n hc x n hc 1 1 2 2 100 \u22c5 \u22c5 = \u22c5 \u03bb \u03bb \u2234 x n n = \u00d7 = \u00d7 = 2 1 1 2 53 100 4530 5080 47 3 \u03bb \u03bb .
Answer: B
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Solution: \u03bb = = 1240 5 248 nm
Answer: A
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Solution: E = n \u22c5 hc \u03c5 = 1 \u00d7 6.626 \u00d7 10 \u201334 \u00d7 3 \u00d7 10 8 \u00d7 1650763.73 = 3.28 \u00d7 10 \u201319 J/quanta.
Answer: C
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Solution: E n h c = \u03bb \u21d2 0 36 6 626 10 3 10 662 6 10 34 8 9 . . . = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 n \u2234 n = 1.2 \u00d7 10 18
Answer: D
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Solution: Energy needed for photochemical dissociation = + = + \u239b \u239d \u239c \u239e \u23a0 \u239f = 482 5 1 2 482 5 96 5 1 2 6 2 . . . . . . KJ mol eV eV eV \u2234 \u03bb \u2248 = 1240 6 2 200 . nm
Answer: A
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Solution: \u03bb \u2248 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1240 289 5 96 5 413 33 . . . nm
Answer: B
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Solution: Energy absorbed per mole of H 2 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u2212 6 10 6 6 10 3 10 270 10 10 440 23 34 8 9 3 . KJ \u2234 Percentage of absorbed energy corrected into K.E. = 440 429 440 100 \u2212 \u00d7 = 2.5 %
Answer: D
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Solution: E = \u00d7 \u00d7 = 9 12400 6900 23 372 kcal/mole \u2234 Energy conversion efficiency = \u00d7 = 111 6 372 100 30 . % EXERCISE II (JEE ADVANCE)\n13.45 Atomic Structure HINTS AND EXPLANATIONS
Answer: C
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Solution: E n hc = \u22c5 \u03bb \u21d2 6 626 6 626 10 3 10 360 10 34 8 9 . . = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 n \u2234 Mole of photons absorbed = = \u00d7 \u00d7 = \u00d7 \u2212 n A ~ . 1 2 10 6 10 2 10 19 23 5 \u2234 Quantum efficiency = \u00d7 \u00d7 \u2212 \u2212 1 10 2 10 0 5 5 5 .
Answer: C
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Answer: C
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Answer: D
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Solution: h h E \u03c5 \u03c5 1 0 = + and h h \u03c5 \u03c5 2 0 = + \u22c5 E K \u2234 \u03c5 \u03c5 \u03c5 0 1 2 1 = \u2212 \u2212 K K
Answer: B
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Answer: C
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Solution: 1 2 2 0 mv hc hc max = \u2212 \u03bb \u03bb \u21d2 v hc m max = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 2 0 0 1 2 \u03bb \u03bb \u03bb \u03bb
Answer: B
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Solution: h h \u03c5 \u03c5 = + K.E. 0 \u21d2 \u03c5 \u03c5 = \u22c5( ) + 1 0 h K.E.
Answer: C
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Solution: c a z \u03bb = \u2212 ( ) 1 and c a z 4 1 \u03bb = \u2032 \u2212 ( ) \u2234 \u2032 = + z z z 1
Answer: B
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Solution: Number of atoms in the disc = \u00d7 \u00d7 = \u00d7 1 12 6 10 5 10 23 22 Now, F K q q r = \u22c5 1 2 2 \u21d2 10 9 10 10 5 9 2 2 2 \u2212 \u2212 = \u00d7 \u00d7 q ( ) \u21d2 q = \u2212 10 3 10 C \u2234 Number of excess electron on negatively charged disc = \u00d7 = \u2212 \u2212 10 3 1 6 10 10 4 8 10 19 9 / . . Hence, Number of excess electron Number of atoms = \u00d7 = \u2212 10 4 8 5 10 10 9 22 / . 1 14 2 4 .
Answer: A
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Solution: r n = r 1 \u00d7 n 2 \u21d2 21.2 \u00d7 10 \u201311 = 5.3 \u00d7 10 \u201311 \u00d7 n 2 \u21d2 n = 2
Answer: C
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Solution: 2 p r n = 26.5 \u00c5 \u21d2 2 p \u00d7 0.529 \u00d7 n 2 2 = 26.5 \u21d2 n = 4
Answer: D
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Solution: r n = 0.529 \u00d7 n z 2 \u00c5
Answer: C
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Solution: r n = r 1 \u00d7 n 2 \u2234 r n \u2013 r n \u2013 1 = r 1 \u00d7 n 2 \u2013 r 1 \u00d7 ( n \u2013 1) 2 = (2 n \u2013 1) \u22c5 r 1 Where n is the higher orbit.
Answer: D
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Solution: A A r r r r r r 2 1 2 2 2 1 2 1 2 1 2 1 2 2 16 1 = = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = ( ) ( ) \u03c0 \u03c0
Answer: C
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Solution: r 4 \u2013 r 2 = 2.116 \u00c5 \u21d2 0.529 \u00d7 4 2 z \u2013 0.529 \u00d7 2 2 z = 2.116 \u2234 z = 3 \u21d2 Li 2+ ion
Answer: A
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Solution: d = 2 p r \u00d7 100 = 2 p \u00d7 0 529 2 4 10 2 10 . \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 m \u00d7 100 = 3.32 \u00d7 10 \u20138 m
Answer: B
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Solution: Circumference = Z p r = 2 p \u22c5 r 0 \u00d7 n 2 1 = 2 p r 0 n 2 and n = 1, 2, 3,.\u2026
Answer: C
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Solution: V n = 2.188 \u00d7 10 6 z n m/s \u21d2 0.547 \u00d7 10 6 = 2.188 \u00d7 10 6 \u00d7 1 n \u2234 n = 4 Now, r n = 0.529 \u00d7 n z 2 = 0.529 \u00d7 4 1 2 = 8.464 \u00c5
Answer: D
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Solution: m r ze r \u03bd \u03c0\u03b5 2 0 2 2 1 4 = \u22c5 \u21d2 \u03bd \u03c0\u03b5 = ( ) ze m r 2 0 4
Answer: D
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Solution: v c ze nh c = ( ) \u22c5 2 4 2 0 \u03c0 \u03c0\u03b5
Answer: A
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Solution: Solution of Q.36
Answer: A
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Solution: F ze r = \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 1 4 9 10 2 1 6 10 4 10 2 88 10 0 2 2 9 19 2 10 2 9 \u03c0\u03b5 ( . ) ( ) . N
Answer: C
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Solution: T n z n z , . sec = \u00d7 \u2212 1 5 10 16 3 2 T T 2 3 3 2 3 2 2 2 3 3 , , / / H Li e + 2+ = \u21d2 T x 2 2 3 , sec H e + =\n13.46 Chapter 13 HINTS AND EXPLANATIONS
Answer: C
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Solution: r r n n 1 2 1 2 2 2 = \u21d2 r r n n 4 2 1 2 2 = \u21d2 n n 1 2 1 2 = \u2234 T T n n 1 1 3 2 3 1 8 2 = =
Answer: C
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Solution: T n a n 3 and n a r n \u21d2 T n a r n 3 2 /
Answer: B
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Solution: N T n z = = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 10 10 1 5 10 2 1 8 33 10 8 8 16 3 2 6 sec , . .
Answer: C
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Solution: \u03bb \u03bd = = \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 c c T n z , . . 3 10 1 5 10 1 1 4 5 10 8 16 3 2 8 m
Answer: A
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Solution: K.E. J V = = \u22c5 ( )\u22c5 = \u22c5 1 2 1 2 2 2 mv mvr r r \u03bd
Answer: C
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Solution: K.E. = = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1 2 1 2 2 8 2 2 2 2 2 2 mv m nh mr n h mr \u03c0 \u03c0 = \u22c5 \u22c5 = \u22c5 n h m a n h ma n 2 2 2 0 2 4 2 2 0 2 2 8 8 \u03c0 \u03c0 ( )
Answer: A
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Solution: \u0394 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = E n n ( ) . . I.E. eV 1 1 14 4 1 1 1 4 13 5 1 2 2 2 2 2
Answer: B
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Solution: Reduced mass effect: \u2032 = \u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f r r m m e n 1 On increasing the nuclear mass, radius decreases.
Answer: C
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Solution: Reduced mass effect: (I.E.) \u2032 = ( )\u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f I.E. 1 1 m m e n On increasing the nuclear mass, ionisation energy increases.
Answer: A
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Solution: u m m m m m m m m m m p p p p p e = + = \u22c5 + = = \u00d7 1 2 1 2 2 1836 2 \u2234 r pm = = 0 529 918 0 058 . . \u00c5
Answer: C
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Solution: n = 2 but z = 3 \u2013 2 = 1 \u2234 E 2 2 2 13 6 1 2 3 4 = \u2212 \u00d7 = \u2212 . . eV and I.E. = 3.4 eV
Answer: A
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Solution: K.E. of emitted electron = 0.5 \u00d7 13.6 eV Now, K.E. mV = 1 2 2 or, 6 8 1 6 10 1 2 9 1 10 19 31 2 . . . \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 \u2212 v \u21d2 v = 1.55 \u00d7 10 6 m/s
Answer: D
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Solution: \u0394 = = = \u00d7 \u2212 E 1240 589 6 2 1 19 . . eV 3.37 10 J = 48.5 kcal/mol
Answer: A
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Solution: \u03bb = = \u00d7 1240 0 0141 8 8 10 4 . . nm = 8.8 \u00d7 10 \u20135 m = 88 nm
Answer: A
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Solution: Minimum is 1 (4 \u2192 1 transition in both atoms) and maximum is 4 (4 \u2192 3 \u2192 2 \u2192 1 in one atom and any other transition in other atom).
Answer: B
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Solution: Number of available orbits is only
Answer: D
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Solution: Hence, maximum number of spectral lines = 4 6 2 C = .
Answer: A
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Solution: 12.1 10.2 1.9 At least two atoms are needed for these three transitions.
Answer: D
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Solution: 1 1 1 2 1 2 1 4 2 1 2 2 2 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f Rz n n R \u21d2 l = 1223 \u00c5 \u2234 UV region.
Answer: C
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Solution: n n ( ) \u2212 = 1 2 15 \u21d2 n = 6 Now, for shortest wavelength, required transition is 6 \u2192
Answer: C
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Solution: \u2234 1 1 1 1 1 6 35 36 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = R R \u21d2 \u03bb = 36 35 R
Answer: C
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Solution: 1 1 1 2 1 4 4 2 2 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2212 R n R n n ( ) \u2234 \u03bb = \u2212 = \u22c5 \u2212 4 4 4 2 2 2 2 n R n K n n ( ) \u21d2 K R = 4
Answer: D
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Solution: \u03bb\u03b1 1 2 z \u21d2 \u03bb \u03bb \u03bb H He Li : : : : : : + + = = 2 1 1 1 2 1 3 36 9 4 2 2 2
Answer: B
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Solution: 1 1 1 1 1 2 2 2 \u03bb = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R n \u21d2 n R R = \u2212 \u03bb \u03bb 1\n13.47 Atomic Structure HINTS AND EXPLANATIONS
Answer: A
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Solution: Required transition is 4 \u2192 2 1 1 1 2 1 4 3 16 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = R R \u21d2 \u03bb = 16 3 R
Answer: C
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Solution: \u03bb\u03b1 1 2 z \u21d2 \u03bb \u03bb Na H 10 1 10 2 2 + = \u21d2 \u03bb Na 10 12 16 + = . \u00c5
Answer: C
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Solution: Excited state is n = 6 [3 \u2192 2, 4 \u2192 2, 5 \u2192 2, 6 \u2192 2] \u2234 Number of spectral lines in 1 R region =
Answer: D
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Solution: 66. Required transition is 3 \u2192
Answer: A
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Solution: Modified Rydberg constant is given by, \u2032 = \u00d7 = R R R me h c 2 2 4 2 4 0 2 3 as \u03c0 \u03c0\u03b5 ( ) Now, 1 1 1 2 1 3 2 5 36 2 2 2 \u03bb = \u2032 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 R R \u2234 \u03bb = 18 5 R
Answer: D
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Solution: \u03c5 = \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 R n R n n 1 2 1 4 4 2 2 2 2 ( )
Answer: B
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Solution: \u0394 = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = E 1312 1 2 1 3 182 22 2 2 . KJ
Answer: A
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Solution: All are visible radiations. Next line is from transition 7 \u2192
Answer: B
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Solution: 70. 410.2 nm n2 n1 2 486.1 nm ? Both are visible radiations. For required series, we get only n 1 . Now, 1 486 1 10 1 09 10 1 1 2 1 9 7 2 2 1 2 . . \u00d7 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 n \u21d2 n 1 =
Answer: A
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Solution: Hence, the series is Brackett series.
Answer: C
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Solution: 1 2 1 2 1 3 1 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R \u21d2 \u03bb 1 9 5 = R 1 2 1 1 1 2 2 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R \u21d2 \u03bb 2 1 5 = R From question, l 1 \u2013 l 2 = 132 nm or, 9 5 1 3 132 10 9 R R \u2212 = \u00d7 \u2212 m \u21d2 R = 1.11 \u00d7 10 9 m \u20131
Answer: D
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Solution: \u221e n + 1 n 1 \u03bb 2 \u03bb \u03c5 = \u00d7 = \u00d7 \u00d7 + \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f 2 725 10 1 09 10 1 1 1 1 6 7 2 2 2 . . ( ) n \u2234 n = 3 Now, 1 1 09 10 2 1 3 1 4 7 2 2 2 \u03bb req = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f . \u21d2 l req = 471.8 nm
Answer: B
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Solution: 1240 108 5 1 2 1 5 2 2 . = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f B.E. \u21d2 B.E. = 54.4 eV
Answer: A
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Solution: K.E. of electron = 13 6 2 1 1 1 2 13 6 27 2 2 2 2 . . . \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = eV Now, 27.2 \u00d7 1.6 \u00d7 10 \u201319 = 1 2 9 1 10 31 2 \u00d7 \u00d7 \u00d7 \u2212 . v \u2234 v \u2248 3.1 \u00d7 10 6 m/s
Answer: D
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Solution: X E n Y E n = = + 2 2 3 and ( ) \u2234 X Y n = + 1 3
Answer: A
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Solution: 1 1 1 1 1 3 2 2 2 \u03bb = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R \u21d2 \u03bb = 9 8 R
Answer: B
\nOriginal PDF solution page
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Solution: n r \u03bb \u03c0 = 2 \u21d2 2 3 3 6 2 \u03c0 \u03c0 \u00d7 \u00d7 = x x
Answer: C
\nOriginal PDF solution page
Open page 664 in PDFExtracted text
Solution: \u03bb = h m 2 E \u21d2 E m \u03b1 \u03bb 1 ( ) for same
Answer: B
\nOriginal PDF solution page
Open page 664 in PDFExtracted text
Solution: \u03bb = h m 2 E \u21d2 \u03bb \u03bb \u03b1 p = \u00d7 \u00d7 = 4 2 1 1 2 2 1
Answer: C
\nOriginal PDF solution page
Open page 664 in PDFExtracted text
Solution: p mv mv v = = = 1 2 1 2 2 const \u21d2 l = Constant
Answer: B
\nOriginal PDF solution page
Open page 664 in PDFExtracted text
Solution: \u03bb min . . = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 1 24 10 5 10 2 48 10 6 4 11 m
Answer: B
\nOriginal PDF solution page
Open page 664 in PDFExtracted text
Solution: m h c h c R = \u22c5 = \u00d7 \u00d7 \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u03bb 2 1 2 1 2 4 10 2 2 2 35 . kg\n13.48 Chapter 13 HINTS AND EXPLANATIONS
Answer: A
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: \u03bb = h m 2 E \u21d2 E E 2 1 1 2 2 2 100 99 1 02 = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2248 \u03bb \u03bb . \u2234 E 2 is about 2 % greater than E 1 .
Answer: C
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: \u0394 = \u22c5 \u0394 = \u22c5 = v h m x h m h mv v min 4 4 4 \u03c0 \u03c0 \u03c0
Answer: C
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: \u03bb = h m 2 E \u21d2 \u0394 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E h m 2 2 2 1 2 2 1 1 \u03bb \u03bb = \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u2212 \u2212 ( . ) . ( ) ( ) 6 626 10 2 9 1 10 1 50 10 1 100 10 34 2 31 9 2 9 2 = 7.24 \u00d7 10 \u201323 J = 4.5 \u00d7 10 \u20134 eV
Answer: C
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: \u03bb = = = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 h m h m 2 2 2 6 626 10 2 4 1 66 10 2 1 6 10 6 34 27 19 E V . . . = 4.15 \u00d7 10 \u201312 m
Answer: A
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: \u03bb = \u22c5 3 32 . n z \u00c5 \u21d2 3 32 3 32 2 . . = \u00d7 n \u21d2 n = 2 Energy of photon liberated in 2 \u2192 1 transition, \u0394 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = E 13 6 2 1 1 1 2 40 8 2 2 2 . . eV \u2234 K.E. of emitted electron from H-atom = 40.8 \u2013 13.6 = 27.2 eV Hence, its de Broglie wavelength is given by, \u03bb = = 150 27 2 2 348 . . \u00c5
Answer: C
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: K.E. of electrons = 12400 3000 12400 4000 1 03 \u2212 = . eV \u2234 \u03bb = = 150 1 03 12 05 . . \u00c5
Answer: D
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: \u0394 \u22c5 \u0394 \u2265 x \u03bb \u03bb \u03c0 2 4 and \u03bb = = 150 6 5\u00c5 \u2234 \u0394 = \u22c5 \u0394 = \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u2212 \u2212 \u03bb \u03bb \u03c0 \u03c0 \u03c0 min ( ) . 2 10 2 9 11 4 5 10 4 1 10 6 25 10 x m
Answer: A
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: \u0394 E = 2.55 eV = 13 6 1 1 1 2 1 2 2 2 . \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f n n eV \u2234 n 1 = 2 and n 2 = 4 Now, \u0394 = \u00d7 \u2212 \u00d7 = \u03bb 3 32 4 1 3 32 2 1 6 64 . . . \u00c5
Answer: B
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: Orbital angular momentum = + \u22c5 l l h ( ) 1 2 \u03c0
Answer: B
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: Electron of 1 s level can never emit photon.
Answer: B
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: Maximum permissible value of l = ( n \u2013 1)
Answer: B
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: m = \u20131 \u21d2 l \u2265 1 \u21d2 can not be s -orbital.
Answer: A
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: Energy 2 s < 2 p < 3 s < 3 p < 4 s < 3 d
Answer: D
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: m = \u20133, \u20132, \u20131, 0, +1, +2, +3
Answer: B
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: Number of radial nodes = n \u2013 l \u2013 1 = 3 \u2013 2 \u2013 1 = 0
Answer: B
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Answer: C
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: Probability of fi nding electron at the nucleus = 0
Answer: D
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Answer: D
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: Mg( z = 12) 1 s 2 2 s 2 2 p 6 3 s 2
Answer: C
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Answer: B
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: 2(1 s ) + 2(2 s ) + 2(2 p ) + 1(3 s ) = 7
Answer: D
\nOriginal PDF solution page
Open page 665 in PDFExtracted text
Solution: L l l h h = + + \u22c5 = \u22c5 ( ) 1 2 5 \u03c0 \u03c0 \u21d2 l = 4 Number of orbitals = 2 l + 1 = 9
Answer: A, B, C
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Solution: (a) (K.E.) Initial = (P.E.) at distance of closest approach or 4.0 MeV = K. q q r 1 2 or, 4 \u00d7 10 6 \u00d7 1.6 \u00d7 10 \u201319 = 9 \u00d7 10 9 \u00d7 ( . ) ( . ) 2 1 6 10 50 1 6 10 19 19 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 r \u2234 Distance of closest approach, r = 3.6 \u00d7 10 \u201314 m (b) P.E. = K \u22c5 q q r 1 2 = 9 \u00d7 10 9 \u00d7 ( . ) ( . ) 2 1 6 10 50 1 6 10 9 10 19 19 14 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 = 10 25 \u00d7 (1.6 \u00d7 10 \u201319 ) 2 J = 10 1 6 10 1 6 10 25 19 2 19 \u00d7 \u00d7 \u00d7 \u2212 \u2212 ( . ) . e V = 1.6 MeV (c) P.E. = K q q r \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 1 2 9 19 19 14 9 10 2 1 6 10 50 1 6 10 4 5 10 1 6 1 ( . ) ( . ) . ( . 0 0 10 19 6 \u2212 \u00d7 ) K q q r \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 1 2 9 19 19 14 9 10 2 1 6 10 50 1 6 10 4 5 10 1 6 1 ( . ) ( . ) . ( . 0 0 10 19 6 \u2212 \u00d7 ) = 3.2 MeV \u2234 K.E. of a -particle at this distance = 4.0 \u2013 3.2 = 0.8 MeV
Answer: A, C, D
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Solution: \u03b5 \u03b5 n n = 1 2
Answer: A, C, D
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Answer: B, C
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Answer: A, C, D
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Answer: A, B, C
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Solution: 1 1 1 1 1028 10 1 09 10 1 1 1 1 2 1 2 2 2 10 7 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 \u00d7 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 RZ n n n . \u2234 n = 3 1 3 2 3 2 1028 \u00c5 1 Induced radiations \u03bb 1 = 1028 \u00c5 \u03bb \u03bb \u03bb 2 1 2 2 2 2 2 1 1 1 3 1 2 1 3 6579 2 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = . \u00c5 \u03bb \u03bb \u03bb 3 1 1 1 3 1 1 1 2 1218 4 1 2 2 2 2 3 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = . \u00c5
Answer: A, B
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Solution: Per atom only one photon is emitted out and hence, the concerned transition is 2 \u2192
Answer: B, C, D
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Solution: \u0394 E Z Z = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = 13 6 1 1 1 2 10 2 2 2 2 2 . . eV
Answer: A, B, C
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Solution: (a) r 1 : r 2 : r 3 = 1 2 : 2 2 : 3 2 = 1 : 4 : 9 (c) \u03bb = = \u00d7 \u00d7 = \u00d7 = \u2212 c v 3 10 6 10 5 10 500 8 14 7 m nm (d) \u03bb \u03bb \u03b1 = \u21d2 h mE m 2 1 \u2234 \u03bb \u03bb \u03bb H : : : : : : H e cn 4 1 1 1 4 1 16 4 2 1 = =
Answer: A, B
\nOriginal PDF solution page
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Solution: h n = \u03d5 + (K.E.) max For A: 4.25 = \u03d5 A + T A and \u03bb A A 2 = h mT For B: 4.20 = \u03d5 B + T B and \u03bb B B = h mT 2 As T B = T A \u2013 1.50 and \u03bb B = 2 \u03bb A \u03d5 A = 2.25 eV; \u03d5 B = 3.70 eV; T A = 2.0 eV; T B = 0.5 eV
Answer: A, C
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Solution: l = 3.32 n z \u00c5
Answer: A, B, C
\nOriginal PDF solution page
Open page 666 in PDFExtracted text
Solution: Theoretical\n13.50 Chapter 13 HINTS AND EXPLANATIONS
Answer: B, C
\nOriginal PDF solution page
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Answer: B, C
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Answer: A, B, C
\nOriginal PDF solution page
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Answer: B, C, D
\nOriginal PDF solution page
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Solution: Na (11) 1 s 2 2 s 2 2 p 6 3 s 1
Answer: A, B, D
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Solution: For minimum l , K.E. of photoelectron should be maximum. For it, the power should be maximum and number of photons is minimum. E max for photon = 5 4 10 1 25 10 18 18 \u00d7 = \u00d7 \u2212 . J \u2234 (K.E.) max of photoelectron = 1.25 \u00d7 10 \u201318 \u2013 4.5 \u00d7 10 \u201319 = 8.0 \u00d7 10 \u201319 J = 5 eV \u2234 \u03bb min = 150 5 30 = \u00c5
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Solution: i min = 4 \u00d7 10 18 \u00d7 1.6 \u00d7 10 \u201319 = 0.64 A
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Solution: i i max min . . = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 9 10 1 6 10 4 10 1 6 10 9 4 18 19 18 19 Comprehension II
Comprehension 1 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Solution: F du dr K r MV r V K mr = \u2212 = = \u21d2 = 4 4 5 2 4 2 (1) From Bohr\u2019s quantization, V n h m r 2 2 2 2 2 2 4 = \u03c0 (2) \u2234 4 4 16 4 4 2 2 2 2 2 2 2 2 K mr n h m r r mK n h nh mK = \u21d2 = = \u03c0 \u03c0 \u03c0 .
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Solution: V nh mr nh m nh mK n h m mK = = = 2 2 4 8 2 2 2 \u03c0 \u03c0 \u03c0 \u03c0 .
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Solution: E = K.E. + P.E. = 1 2 2 4 mv K r + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2212 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 2 4 4 4 4 4 m K mr K r K nh mK \u03c0 \u2234 E n h m K = 4 4 4 2 256 \u03c0 Comprehension III \u0394 E z n n = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 13 6 1 1 2 2 2 1 2 . eV 10.2 + 17.0 = 13.6 z 2 1 2 1 2 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f n (1) 4.25 + 5.95 = 13.6 z 2 1 3 1 2 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f n (2)
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Comprehension 2 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 667 in PDFExtracted text
Solution: \u0394 E = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = 13 6 3 1 1 1 7 119 9 2 2 2 . . eV\n13.51 Atomic Structure HINTS AND EXPLANATIONS Comprehension IV
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: After excitation, n =
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: Hence, initial excited state is n =
Comprehension 3 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: 1 1 1 1 1654 10 1 09 10 1 2 1 3 2 1 2 2 2 10 7 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 \u00d7 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 RZ n n z . \u2234 z = 2 \u00de He + ion
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: \u0394 E z n n = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f = 13 6 1 1 13 6 2 1 3 1 6 04 2 1 2 2 2 2 2 2 . . . eV Comprehension V
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: Final excited state, after absorption of 2.7 eV, is
Comprehension 4 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: On de-excitation, the sample emit radiations equal to less than or more than 2.7 eV and hence, the initial excited state must be
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: 4 2.7 eV 2.7 eV Less than 2.7 eV More than 2.7 eV 3 2 1
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: \u0394 E I E n n I E = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f . . . . . 1 1 2 7 1 2 1 4 1 2 2 2 2 2 \u2234 I . E . = 14.4 eV
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: \u0394 E I E n n min . . . = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 1 14 4 1 3 1 4 1 2 2 2 2 2 = 0.7 eV Comprehension VI
Comprehension 5 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: r n h mze n z = = \u00d7 ( ) . 4 4 0 529 0 2 2 2 2 2 \u03c0\u03b5 \u03c0 \u00c5 (for H-like atom) For this system, r = \u00d7 = \u00d7 = \u2212 0 529 1 207 2 56 10 0 256 3 . . . \u00c5 pm
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: I . E . = 2 4 13 6 2 2 4 0 2 2 2 2 2 \u03c0 \u03c0\u03b5 mz e n h z n ( ) . = \u00d7 eV ( for H-like atom) For this system, I . E . = 13.6 \u00d7 207 = 2835.9 eV
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: Rydberg constant for this system = 1.09 \u00d7 10 7 \u00d7 207 m \u20131 \u2234 1 1 09 10 207 1 1 1 2 5 91 10 7 2 2 10 \u03bb \u03bb = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u00d7 \u2212 ( . ) . m Comprehension VII
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: n n n ( ) \u2212 = \u21d2 = 1 2 6 4 Now, 1 1 1 1 10 1 09 10 1 1 1 1 4 2 1 2 2 2 10 7 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 \u00d7 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 RZ n n x . \u2234 x = 978.6
Comprehension 6 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 668 in PDFExtracted text
Solution: For max l , transition : n = 4 to n = 3 \u2234 1 1 09 10 1 1 3 1 4 1 887 10 7 2 2 2 6 \u03bb \u03bb max max . . = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u00d7 \u2212 m\n13.52 Chapter 13 HINTS AND EXPLANATIONS
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: For max n , transition : n = 4 to n = 1 \u2234 Hz \u03bd \u03bb max . . = = \u00d7 \u00d7 = \u00d7 \u2212 c 3 10 978 6 10 3 066 10 8 10 15
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: 1R radiations involve transition : n = 4 to n = 3 only.
Comprehension 7 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: Visible radiation involve transitions : n = 4 to n = 2 (489.3 nm) and n = 3 to n = 2 (660.5 nm). Comprehension VIII
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: 5 4 3 2 1 1 2 3 4 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: 5 3 2 1 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af and 5 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af 4 3 1
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: 6 + 1 (any possibility after than Q.27)
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: 5 5 1 2 10 ( ) \u2212 =
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: 4 5 3 2 1 1 1 1 1 3 4 2 5 6 2 Comprehension IX \u0394 E Z n n eV Z n = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 13 6 1 1 204 13 6 1 1 1 2 2 1 2 2 2 2 2 2 . . ( ) (1) 40 8 13 6 1 1 2 2 2 2 . . ( ) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f Z n n (2)
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: n = 2 \u21d2 2 n = 4
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: E 1 2 2 13 6 4 1 217 6 = \u2212 \u00d7 = \u2212 . . eV
Comprehension 10 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: \u0394 E min . . = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = 13 6 4 1 3 1 4 10 58 2 2 2 eV Comprehension X
Comprehension 11 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: 1 1 1 2 1 4 1 4 16 16 2 1 2 2 2 2 2 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 R Z n n R n R n n ( ) \u2234 \u03bb = \u2212 = \u2212 \u21d2 = = 4 16 16 4 366 97 2 2 2 2 n R n cn n C R ( ) . nm
Comprehension 11 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: For series limit, n = \u221e \u2234 l = C = 366.97 nm
Comprehension 11 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: For n = 5, l = 1019.36 nm For n = \u221e , l = 366.97 nm Comprehension XI
Comprehension 12 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Comprehension 12 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: E E S H 1 13 6 3 2 2 25 2 2 = \u2212 \u00d7 = \u00d7 . .
Comprehension 12 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 669 in PDFExtracted text
Solution: S 2 = 3 p \u21d2 l = 1\n13.53 Atomic Structure HINTS AND EXPLANATIONS Comprehension XII
Comprehension 13 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: n = 2 l = 1 j = 3 2 1 2 or m = = \u2212 \u2212 + \u2212 + \u2212 3 2 1 2 1 2 3 2 , , , for j = 3 2 = \u2212 + 1 2 1 2 , for j = 1 2
Comprehension 13 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: n = 3 l = 0 \u21d2 j = 1 2 \u21d2 m = \u2212 + 1 2 1 2 , = 2 \u21d2 j = 3 2 \u21d2 m = \u2212 \u2212 + + 3 2 1 2 1 2 3 2 , , , = \u21d2 = \u2212 \u2212 \u2212 + + + 5 2 5 2 3 2 1 2 1 2 3 2 5 2 m , , , , , Comprehension XIII
Comprehension 13 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: Radial nodes = n \u2013 l \u2013 1 = 3 Angular nodes = 1
Answer: A
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: Orbital angular momentum = + \u22c5 l l h ( ) 1 2 \u03c0
Answer: B
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Answer: D
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Answer: C
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: Spin quantum number is independent from wave function.
Answer: D
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Answer: C
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: Be is the reactive element.
Answer: A
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: 2 p \u21d2 2 p x + 2 p y + 2 p z \u21d2 Total 3 angular nodes.
Answer: D
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: dz 2 has two conical nodes.
Answer: D
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: 3 p x and 3 p y diff ers in angular function.
Answer: C
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: 4s energy level is lower than 3 d .
Answer: A \u2192 R, S; B \u2192 Q, S; C \u2192 P, Q; D \u2192 P, R
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: Radial nodes = n \u2013 l \u2013 1 Angular nodes = l
Answer: A \u2192 R, S; B \u2192 P, S; C \u2192 Q; D \u2192 Q
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Answer: A \u2192 R; B \u2192 Q; C \u2192 P; D \u2192 S
\nOriginal PDF solution page
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Solution: (A) V K P E K E n n = = \u2212 = \u2212 . . . . mV mV 2 1 2 2 2 (B) \u03b5 n n r \u221d \u2212 ( ) 1 (C) Lowest energy level is 1 s . (D) r z n \u221d 1
Answer: A \u2192 S; B \u2192 P; C \u2192 R; D \u2192 Q
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Answer: A \u2192 P; B \u2192 S; C \u2192 Q; D \u2192 R
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: Graph of s -orbital status with some value but for other orbitals, it starts from zero. Radial nodes: 3 s = 2, 4 s = 3, 2 p = 0, 3 p = 1
Answer: A \u2192 P, U; B \u2192 Q, T; C \u2192 S, W; D \u2192 R, V
\nOriginal PDF solution page
Open page 670 in PDFExtracted text
Solution: (A) r n z \u221d 2 (B) V z n \u221d (C) F mv r z n = \u221d 2 3 4 (D) f v r z n = \u221d 2 2 3 \u03c0\n13.54 Chapter 13 HINTS AND EXPLANATIONS
Answer: A \u2192 P; B \u2192 P, Q, S; C \u2192 P, R; D \u2192 Q, S
\nOriginal PDF solution page
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Solution: (A) 3 radial nodes \u21d2 4 s , 5 p , 6 d but graph does not start from origin and hence, only 4 s . (B) 3 radial nodes \u21d2 4 s , 5 p , 6 d (C) Only s- orbital (D) l \u2265 1
Answer: A \u2192 Q, R; B \u2192 P, Q, R, S; C \u2192 P, Q, R; D \u2192 P, Q
\nOriginal PDF solution page
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Answer: A \u2192 R, S; B \u2192 Q, P; C \u2192 P
\nOriginal PDF solution page
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Solution: 10. (A) V V 6 4 4 6 2 3 = = v \u221d \u239b \u239d \u239c \u239e \u23a0 \u239f 1 n (B) \u03bb \u03bb 3 2 1 1 1 1 2 1 4 1 2 2 2 2 = \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f = (C) \u03bb \u03bb c p = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1 3 1 6 1 1 1 3 3 3 2 2 2 2 2 (D) \u0394 \u0394 E E n H e + = = 1 2 1 4 2 2
Answer: A \u2192 Q; B \u2192 P; C \u2192 S; D \u2192 R
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+ "title": "Chem Sec 6",
+ "originalName": "Section F - Integer",
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+ {
+ "question_id": "atomic-structure-chem-sec-6-1-164",
+ "marks": 4.0,
+ "negMarks": 1.0,
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+ "subject": "chemistry",
+ "chapter": "atomic-structure",
+ "chapterTitle": "Atomic Structure",
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+ "originalNumber": 164,
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\nOriginal PDF solution page
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Solution: \u03b5 = = 1240 300 4 13 . eV For photoelectric effect, e \u2265 f \u21d2 N 0 = 4
Answer: 6
\nOriginal PDF solution page
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Solution: Frequency of reduction \u221d z n 2 3 \u2234 T T 3 2 2 3 8 2 2 7 1 3 4 8 10 1 2 1 28 10 1 6 = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 . .
Answer: 5
\nOriginal PDF solution page
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Solution: 1 1 1 2 1 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R Z n n \u21d2 1 108 5 10 1 30 4 10 1 09 10 2 1 1 1 7 7 7 2 2 2 . . . \u00d7 + \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 n \u2234 n = 5
Answer: 3
\nOriginal PDF solution page
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Solution: \u03bb 3 \u2013 \u03bb 2 = 59.3 nm or, 1 1 2 1 3 1 1 1 1 2 59 3 2 2 2 2 2 2 R Z R Z \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = . nm \u21d2 z = 3
Answer: 3
\nOriginal PDF solution page
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Solution: Final excited state = 5th orbit As only the wavelengths are longer than absorbed radiation initial excited state = 3rd orbit
Answer: 4
\nOriginal PDF solution page
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Solution: \u0394 E = 12.75 = 13.6 \u00d7 1 2 4 2 1 4 2 n m n n \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 =
Answer: 5
\nOriginal PDF solution page
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Solution: \u0394 \u0394 x h m V min . . = = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 4 6 626 10 4 10 3 313 10 5 10 34 6 3 26 \u03c0 \u03c0 \u03c0 m
Answer: 1
\nOriginal PDF solution page
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Solution: m h x v min . . = \u22c5 \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u2212 4 6 626 10 4 10 5 27 10 1 34 11 24 \u03c0 \u03c0 \u0394 \u0394 kg
Answer: 2
\nOriginal PDF solution page
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Solution: 2 p r = nl \u21d2 l = 4 2 nm = 2 nm
Answer: 2
\nOriginal PDF solution page
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Solution: For radial node, y 23 = 0 \u21d2 r 0 = 2 a 0 Four-digit Integer Type
Answer: 0025
\nOriginal PDF solution page
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Solution: Initial K. E. = P. E. at distance of closest approach or, p q q r 2 0 1 0 2 1 4 m = \u22c5 \u03c0\u03b5 . or ( . ) . . . 3 2 10 2 4 10 6 10 9 10 2 1 6 10 1 6 10 1 20 2 3 23 9 19 19 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 z 5 5 10 13 \u00d7 \u2212 \u2234 z = 25 Orbital Radial nodes Angular nodes 3 d 0 2 2 p 0 1 3 p 1 1 5 d 2 2\n13.55 Atomic Structure HINTS AND EXPLANATIONS
Answer: 0084
\nOriginal PDF solution page
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Solution: t = = \u00d7 \u00d7 \u00d7 = Distance Speed sec 2 12600 10 3 10 0 084 3 8 .
Answer: 0030
\nOriginal PDF solution page
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Solution: c a \u03bb = \u2212 (z ) 6 c a 180 27 1 = \u2212 ( ) c a z z 144 1 30 = \u2212 \u21d2 = ( )
Answer: 8400
\nOriginal PDF solution page
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Solution: nh n = ms \u2219 \u0394 T or, n \u00d7 6.626 \u00d7 10 \u201334 \u00d7 2.45 \u00d7 10 10 = 245 \u00d7 4.2 \u00d7 (99.5 \u2013 19.5) \u2234 Number of photons = 5.04 \u00d7 10 27 \u2234 Moles of photon = 5 04 10 6 10 8400 27 23 . \u00d7 \u00d7 =
Answer: 0917
\nOriginal PDF solution page
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Solution: \u0394 E = \u0394 E 1 + \u0394 E 2 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 + \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 1310 1 1 1 3 45 100 1310 1 1 1 2 40 100 917 2 2 2 2 kJ
Answer: 0091
\nOriginal PDF solution page
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Solution: \u03bb = = 1240 13 6 91 17 . . nm
Answer: 0100
\nOriginal PDF solution page
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Solution: Moles of H 2 = PV RT x = \u00d7 \u00d7 = 1 1 0 08 300 . \u0394 E x x = \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 436 1312 1 1 1 2 2 100 16 2 2 . kJ
Answer: 0090
\nOriginal PDF solution page
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Solution: 2 1 ? 360 nm 120 nm \u221e 1 1 120 1 360 90 \u03bb \u03bb = + \u21d2 = nm
Answer: 0024
\nOriginal PDF solution page
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Solution: \u03bb = \u21d2 = \u21d2 = 150 2 5 150 1 2 V V 24 \u00af V V .
Answer: 0005
\nOriginal PDF solution page
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Solution: \u03bb \u03bb\u03b1 = = \u22c5 \u21d2 \u22c5 h h m KT m T 2 2 3 2 1 mE \u2234 \u03bb \u03bb H N e e = \u00d7 \u00d7 = 20 1000 4 200 5
![\"Chemical](\"./assets/chapter-packs/chemical-equilibrium/Chemistry)
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+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: K eq for the reaction in backward direction = = \u00d7 \u00d7 = \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 K K s b f 2 1 10 3 9 10 53 846 3 1 1 5 1 1 . . . L mol s L mol
Answer: C
\nOriginal PDF solution page
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Solution: Stability constant, K K K f b = = \u00d7 \u00d7 = \u00d7 \u2212 1 45 10 1 22 10 1 1885 10 13 9 17 . . .
Answer: C
\nOriginal PDF solution page
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Solution: K K K B A f b eq = = [ ] [ ] 2 \u21d2 1 5 10 100 10 10 10 3 2 5 . ( / ) ( / ) \u00d7 = \u2212 \u2212 K b \u21d2 K b = 1.5 \u00d7 10 \u201311 M \u20131 S \u20131
Answer: A
\nOriginal PDF solution page
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Solution: For pent hydrate to be efflorescent, Q < K p or, P H O 2 atm 2 4 2 10 < \u2212 \u21d2 P H O 2 atm mm < = \u2212 10 7 6 2 .
Answer: C
\nOriginal PDF solution page
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Solution: Q P P K P P = \u00d7 = \u00d7 = < NH 2 CO 3 atm 2 10 20 2000 2 3 Hence, the reaction should shift forward. But as solid NH 2 COONH 4 is not present initially, the pressure will remain at 30 atm.
Answer: D
\nOriginal PDF solution page
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Solution: Q K P K P P p < \u21d2 < H O 2 2 \u21d2 40 760 100 1 21 10 2 4 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f < \u00d7 \u2212 R H . . \u2234 R.H. < 20.9 %
Answer: B
\nOriginal PDF solution page
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Solution: H 2 (g) + I 2 (s) \u001f 2HI(g) ; K p = 6.4 \u00d7 10 \u20134 atm I 2 (s) \u001f I 2 (g) ; K p = 1.6 \u00d7 10 \u20134 atm \u2234 H 2 (g) + I 2 (g) \u001f 2HI(g) ; K P = \u00d7 \u00d7 = \u2212 \u2212 6 4 10 1 6 10 4 4 4 . .
Answer: A
\nOriginal PDF solution page
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Solution: Net rate of reaction of HI, \u2212 \u22c5 = \u2212 = \u2212 \u2212 1 2 1 2 1 2 2 d dt r r b f [ ] [ ] [ ][I ] HI K HI K H
Answer: D
\nOriginal PDF solution page
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Solution: \u03b1 = \u2212 \u2212 \u22c5 = \u2212 \u2212 \u00d7 = M M n M 0 1 208 5 124 2 1 124 0 681 ( ) . ( ) .
Answer: A
\nOriginal PDF solution page
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Solution: PCl g PCl g 3 2 5 ( ) Cl ( ) ( ) + g \u001f \u21c0 \u001f \u21bd \u001f \u001f Initial partial pressure P 0 P 0 0 Equilibrium partial pressure P 0 \u2013 0.75 P 0 P 0 \u2013 0.75 P 0 0.75 P 0 \u20130.25 P 0 \u20130.25 P 0 Now, K P P P P = \u00d7 PCl PCl Cl 5 3 2 \u21d2 2 0 75 0 25 0 25 0 0 0 = \u00d7 . . . P P P \u21d2 P 0 = 6 atm \u2234 Initial total pressure of mixture = 2 P 0 = 12 atm
Answer: A
\nOriginal PDF solution page
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Solution: N g 3H g NH g 2 2 3 2 ( ) ( ) ( ) + \u001e \u21c0 \u001e \u21bd \u001e \u001e Initial moles 1 3 0 Moles at equilibrium 1 \u2013 x 3 \u2013 3 x 2 x Total moles of gases = (1 \u2013 x ) + (3 \u2013 3 x ) + 2 x = 4 \u2013 2 x Equilibrium partial pressure 1 4 2 3 3 4 2 2 4 2 \u2212 \u2212 \u00d7 \u2212 \u2212 \u00d7 \u2212 \u00d7 x x P x x P x x P Now, K x x P x x P x x P x P = \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 2 4 2 1 4 2 3 3 4 2 4 27 1 2 3 2 ( \u2212 \u2212 \u00d7 \u2212 \u2248 \u00d7 \u00d7 x x P x P ) ( ) 4 2 2 2 2 4 2 4 16 27 \u2234 x K P P K P P = \u22c5 = \u22c5 27 64 3 3 8 2
Answer: D
\nOriginal PDF solution page
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Solution: X 2 + Y 2 \u001f 2 XY Initial moles 2 3 0 Final moles 2 \u2013 x 3 \u2013 x 2 x\n6.37 Chemical Equilibrium HINTS AND EXPLANATIONS [ ] . XY = = 2 5 0 7 x \u21d2 x = 1.75 \u2234 [X ] . 2 2 5 0 05 = \u2212 = x M and [Y ] . 2 3 5 0 25 = \u2212 = x M
Answer: A
\nOriginal PDF solution page
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Solution: N 2 + O 2 \u001f 2NO Equilibrium moles 1 \u2013 x 1 \u2013 x 2 x 0 09 2 1 1 2 . ( ) ( )( ) = \u2212 \u2212 x x x \u21d2 x = 0.13
Answer: B
\nOriginal PDF solution page
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Solution: N 2 + O 2 \u001f 2NO Initial moles 4 a a 0 Equilibrium moles 4 a \u2013 x a \u2013 x 2 x Now, 0 0004 2 4 4 4 2 2 . ( ) ( )( ) = \u2212 \u2212 \u2248 \u22c5 x a x a x x a a \u21d2 x a = 0 02 . \u2234 Per cent of NO = 2 5 100 0 8 x a \u00d7 = . %
Answer: A
\nOriginal PDF solution page
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Solution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 1 5 0 Moles at equilibrium 1 \u2013 x 5 \u2013 3 x 2 x Total moles = (1 \u2013 x ) + (5 \u2013 3 x ) + 2 x = 6 \u2013 2 x From question, 2 6 2 0 4 x x \u2212 = . \u21d2 x = 6 7 K x x x x P P = \u2212 \u00d7 \u2212 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u2212 ( ) ( ) ( ) . 2 1 5 3 6 2 2 6 10 2 3 2 4 2 atm
Answer: B
\nOriginal PDF solution page
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Solution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 4 16 0 Moles at equilibrium 4 \u2013 x 16 \u2013 3 x 2 x Total moles = (4 \u2013 x ) + (16 \u2013 3 x ) + 2 x = 20 \u2013 2 x From question, 20 9 10 20 2 \u00d7 =\u2212 x \u21d2 x = 1 Now, K x x x V C = \u2212 \u2212 \u22c5 = \u00d7 \u2212 \u2212 ( ) ( )( ) . 2 4 16 3 6 07 10 2 3 2 4 2 M
Answer: B
\nOriginal PDF solution page
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Solution: If reactants are taken in stoichiometric amount, then their mass ratio does not change at any stage of reaction. For 3 mole N 2 , there should be 9 mole H 2 . Hence, at any stage, m m m N H NH 2 3 gm 2 3 28 9 2 102 + + = \u00d7 + \u00d7 = .
Answer: A
\nOriginal PDF solution page
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Solution: 2SO 2 + O 2 \u001f 2SO 3 Initial moles 2 1 0 Moles at equilibrium 2 \u2013 2 x 1\u2013 x 2 x From question n eq SO 2 = n eq MnO 4 \u2212 . or, (2 \u2013 2 x ) \u00d7 2 = 0.4 \u00d7 5 \u21d2 x = 0.5 \u2234 K x x x C = \u2212 \u00d7 \u2212 = \u2212 ( ) ( ) ( ) 2 2 2 1 2 2 2 1 M .
Answer: D
\nOriginal PDF solution page
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Solution: CH 3 COOH + C 2 H 5 OH \u001f CH 3 C00C 2 H 5 + H 2 O Case I 60 60 1 = mole 46 46 1 = mole 0 0 Moles at Equ. 1 \u2013 x 1 \u2013 x x = = 44 88 0 5 . x Case II 120 60 2 = mole 46 46 1 = mole 0 0 Moles at Equ. 2 \u2013 y 1 \u2013 y y y K x x x x y y y y eq = \u22c5 \u2212 \u22c5 \u2212 = \u22c5 \u2212 \u22c5 \u2212 ( ) ( ) ( ) ( ) 1 1 2 1 \u21d2 y = 2 3 \u2234 Mass of CH 3 COOC 2 H 5 at equilibrium = 2 3 88 \u00d7 =
Answer: D
\nOriginal PDF solution page
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Solution: R 1 OH + CH 3 COOH \u001f CH 3 COOR 1 + H 2 O Initial moles 1 1 0 0 Equ. moles 1 \u2013 x 1 \u2013 ( x + y ) x x + y\n6.38 Chapter 6 HINTS AND EXPLANATIONS R 2 OH + CH 3 COOH \u001f CH 3 COOR 2 + H 2 O Initial moles 1 1 0 0 Equ. moles 1 \u2013 y 1 \u2013 ( x + y ) y x + y From question, x + y = 0.8 and x y = 3 2 \u2234 x = 0.48 and y = 0.32 Now, K x x y x x y 1 1 1 0 48 0 8 0 52 0 2 3 69 = \u22c5 + \u2212 \u2212 + = \u00d7 \u00d7 = ( ) ( )[ ( )] . . . . .
Answer: B
\nOriginal PDF solution page
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Solution: 2NO( g ) + Cl 2 ( g ) \u001f 2NOCl( g ) Initial partial pressure 2P 0 P 0 0 Equ. partial pressure 2P 0 \u2013 2x P 0 \u2013 x 2x From question, (2 P 0 \u2013 2 x ) + ( P 0 \u2013 x ) + 2 x = 1 \u21d2 3 P 0 \u2013 x = 1 (1) and 2 1 4 0 x P x = \u2212 ( ) (2) From (1) and (2), P 0 = 9 x and x = 1 26 \u2234 K x P x P x P = \u2212 \u2212 = \u2212 ( ) ( ) ( ) 2 2 2 13 256 2 0 2 0 1 atm
Answer: B
\nOriginal PDF solution page
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Solution: S 8 ( g ) \u001f 4 S 2 ( g ) Initial partial pressure 1 atm 0 Equ. partial pressure 1 \u2013 0.3 4 \u00d7 0.3 = 0.7 atm = 1 .2 atm \u2234 K P = = ( . ) . . 1 2 0 7 2 96 4 3 atm
Answer: D
\nOriginal PDF solution page
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Solution: HCl( ) O Cl H O 2 g g g g + + 1 4 1 2 1 2 2 2 ( ) ( ) ( ) \u001e \u21c0 \u001e \u21bd \u001e \u001e Initial partial pressure 730 8 100 \u00d7 730 92 100 \u00d7 = 58.4 mm = 671.6 mm Equilibrium partial pressure 58.4 \u2013 58.4 \u00d7 0.08 671 6 58 4 0 08 4 . . . \u2212 \u00d7 = 670.432 mm
Answer: B
\nOriginal PDF solution page
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Solution: H 3 BO 3 + Glycerin \u001f complex Initial concent. 0.1 a M 0 Equ. Concert 0.1 \u2013 0.06 ( a \u2013 0.06) M 0.06 M = 0.04 M Now, K a eq = = \u00d7 \u2212 0 9 0 06 0 04 0 06 . . ( . ) ( . ) \u21d2 a = 1.73 M
Answer: B
\nOriginal PDF solution page
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Solution: 2A( g ) \u001f A 2 ( g ); K P = 8 \u00d7 10 8 atm \u20131 Initial partial pressure 1 atm 0 Partial pressure on complete reaction 0 0.5 atm Equilibrium partial pressure 2 x atm 0.5 \u2013 x \u2248 0.5 atm Now, 8 10 0 5 8 2 \u00d7 = . P A \u21d2 P A = 2.5 \u00d7 10 \u20135 atm
Answer: C
\nOriginal PDF solution page
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Solution: K eq = 3.8 \u00d7 10 \u20137 10 6 3 2 \u2212 \u2212 \u00d7 [ ] [ ] HCO CO \u21d2 [ ] [ ] . HCO CO 3 2 0 38 \u2212 =
Answer: A
\nOriginal PDF solution page
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Solution: A(g) \u001f nB(g) Initial mole 1(say) 0 Equilibrium mole 1 \u2013 a n a Total moles = 1 \u2013 a + n a = 1 + a ( n \u2013 1) Now K P P n n P n P n P P B n A n n n = = + \u2212 \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 + \u2212 \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u22c5 \u2212 \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 1 1 1 1 1 1 ( ) ( ) ( ) ( ( ) [ ( )] 1 1 1 1 \u2212 \u22c5 + \u2212 \u2212 \u03b1 \u03b1 n n
Answer: B
\nOriginal PDF solution page
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Solution: A + B \u001f C + D Initial moles a a 0 0 Equilibrium moles a \u2013 x a \u2013 x x x From question, [ A ] = 2[ C ] \u21d2 a \u2013 x = 2 x \u21d2 a = 3 x Now, K K K x x a x a x f b eq = = \u22c5 \u2212 \u22c5 \u2212 ( ) ( ) \u21d2 2 10 2 2 3 \u00d7 = \u22c5 \u22c5 \u2212 K x x x x b \u2234 K b = 8 \u00d7 10 \u20133 mol \u20131 L S \u20131\n6.39 Chemical Equilibrium HINTS AND EXPLANATIONS
Answer: D
\nOriginal PDF solution page
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Solution: (Cl 2 CHCOOH) 2 \u001f 2Cl 2 CHOOH Initial Conc. 0 0129 258 100 1000 . / / 0 = 5 \u00d7 10 \u20134 M Equ. Conc. 5 \u00d7 10 \u20134 \u2013 x 2 x Now, K eq = 5 \u00d7 10 \u20134 = ( ) ( ) 2 5 10 2 4 x x \u00d7 \u2212 \u2212 \u21d2 x = 1.95 \u00d7 10 \u20134 \u2234 [Cl 2 CHOOH] = 3.90 \u00d7 10 \u20134 M
Answer: C
\nOriginal PDF solution page
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Solution: NH 2 CONH 4 ( s ) \u001f 2NH 3 ( g ) + CO 2 ( g ) Initial moles 1 0 0 Equ. moles 1 \u2013 a 2 a a From question, 3 \u03b1 = \u22c5 P V RT \u2234 Percentage dissociation of solid = 100 a % = \u22c5 100 3 PV RT %
Answer: A
\nOriginal PDF solution page
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Solution: P H O eq 2 , = ( K P ) 1/4 = (8.1 \u00d7 10 \u20137 ) = 0.03 atm P H O eq actual 0.04 atm 2 30 4 760 , , . = = \u2234 Mass of water vapour absorbed = ( . . ) . . 0 09 0 03 1 642 0 0821 300 18 \u2212 \u00d7 \u00d7 \u00d7 = 0.012 gm
Answer: C
\nOriginal PDF solution page
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Solution: Addition of CO will shift second reaction backward. Decrease in Cl 2 will shift the fi rst reaction forward.
Answer: B
\nOriginal PDF solution page
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Solution: P g P P g P H O HCl(g) H O HCl(g) 2 2 new ( ) ( ) , 2 2 2 = \u00d7 \u21d2 P P HCl (g), new HCl (g) = \u00d7 2
Answer: C
\nOriginal PDF solution page
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Solution: Ionic form of the reaction is NH H O NH OH H 4 2 4 + + + + \u001e \u21c0 \u001e \u21bd \u001e \u001e
Answer: D
\nOriginal PDF solution page
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Solution: K A B AB K AB AB B 1 2 2 = = + \u2212 \u2212 \u2212 [ ][ ] [ ] [ ] [ ][ ] and Now, [ ] [ ] [ ] A AB K K B + \u2212 \u2212 = \u22c5 2 1 2 2
Answer: B
\nOriginal PDF solution page
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Solution: NH 4 HS(s) \u001f NH 3 ( g ) + H 2 S( g ) Equ. partial pressure P 2 atm P 2 atm New Equ.partial pressure P atm P \u2032 atm Now, P P P P 2 2 \u00d7 = \u00d7 \u2032 \u21d2 P \u2032 = 0.25 P
Answer: A
\nOriginal PDF solution page
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Solution: N 2 + 3H 2 \u001f 2NH 3 Equ. partial pressure 100 mm 400 mm 1000 mm New equ. partial pressure 100 \u2013 a + x 400 + 3 x = 700 mm 1000 \u2013 2 x = 800 mm \u2234 x = 100 mm Now, K P P N = \u00d7 = \u00d7 1000 100 400 800 700 2 3 2 3 2 \u21d2 P N 2 11 94 = . mm K B A K C A 1 2 = = [ ] [ ] , [ ] [ ] Now, X A A B C A A K A K A K K A = + + = + + = + + [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 1 2 1 2 1 1
Answer: B
\nOriginal PDF solution page
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Solution: CO and H 2 are initially in 1 : 3 mole ratio, as they are formed by 2nd reaction. CO + 2H 2 \u001f CH 3 OH Initial moles 1 3 0 Equilibrium moles 1 \u2013 0.25 3 \u2013 0.25 \u00d7 2 0.25 = 0.75 = 2.5 Total moles = 0.75 + 2.5 + 0.25 = 3.5 Now, K P P = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u2212 0 25 0 75 2 5 3 5 6 23 10 2 2 3 . . ( . ) . . \u2234 P = 10.24 bar
Answer: C
\nOriginal PDF solution page
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Solution: A \u001f B + C; K 1 = 10 6 Initial moles 1 0 0 Equilibrium moles 1 \u2013 x + y x \u2013 y x B + D \u001f A; K 2 = 10 \u20136 x 1 1 Equilibrium moles x \u2013 y 1 \u2013 y 1 + y \u2013 x\n6.40 Chapter 6 HINTS AND EXPLANATIONS As K 1 >> 1, we may assume x \u2248 1 Now, K y x x y y y y y 2 1 1 1 1 = + \u2212 \u2212 \u2212 \u2248 \u2212 \u22c5 \u2212 ( ) ( )( ) ( ) ( ) As K 2 << 1, we may assume y << 1 K y y y y 2 1 1 = \u2212 \u2212 ( )( ) \u001a \u2234 [A] = 1 \u2013 x + y \u2248 y = 10 \u20136 M
Answer: A
\nOriginal PDF solution page
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Solution: A \u001f B Initial a M b M Equilibrium ( a \u2013 x ) M ( b + x ) M Now, K K K b x a x eq = = + \u2212 1 2 \u2234 x K a K b K K = \u2212 + 1 2 1 2
Answer: A
\nOriginal PDF solution page
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Solution: r b = K b \u22c5 P C(g)
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 G \u00b0 = \u20132.303 RT \u22c5 ln K p \u00b0 \u0394 H \u00b0 \u2013 T \u22c5 \u0394 S \u00b0 or \u20132.303 \u00d7 8.314 \u00d7 T \u00d7 ln 1.0 = 240 \u00d7 10 3 \u2013 T \u00d7 50 \u2234 T = 4800 K
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 G \u00b0 = \u20132.303 RT \u22c5 ln K eq \u21d2 \u20132.303 \u00d7 10 3 = \u20132.303 \u00d7 2 \u00d7 500 \u00d7 ln K eq \u2234 K eq = 10 Now, K P P P eq = \u00d7 HI H I 2 2 1 2 1 2 / / \u21d2 10 0 001 2 1 2 1 2 P H / / ( . ) \u00d7 \u2234 P H atm 2 1000 =
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 \u00b0 = \u2212 \u22c5 = \u2212 \u22c5 = \u2212 \u22c5 G RT B RT RT ln [ ] [ ] ln ln . \u03b1 64 36 1 78
Answer: C
\nOriginal PDF solution page
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Solution: There is no net change at equilibrium.
Answer: D
\nOriginal PDF solution page
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Solution: K eq at 27\u00b0C, K 1 3 4 2 10 2 10 4 = \u00d7 \u00d7 = \u2212 \u2212 and K eq at 127\u00b0C, K 2 2 3 8 10 4 10 20 = \u00d7 \u00d7 = \u2212 \u2212 Now, ln K K H R T T 2 1 1 2 1 1 = \u0394 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, 2 303 20 4 1 300 1 400 . log = \u0394 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f H R \u2234 \u0394 H = 2.303 \u00d7 8.314 \u00d7 1200 log(5) J/mol
Answer: D
\nOriginal PDF solution page
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Solution: n A \u001f A n Initial moles 1 0 Equilibrium moles 1 \u2013 x x n Now, K x n V x V x V x n x V n x C n n n n = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2212 \u22c5 \u2248 \u22c5 << \u2212 \u2212 / ( ) ( ) 1 1 1 1 1 as Now, total moles = (1 \u2013 x ) + x n = 1 + x \u22c5 1 1 n \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u2212 \u22c5 \u2212 \u2212 1 1 1 1 1 1 n KC V n n n K V n C n ( )
Answer: A
\nOriginal PDF solution page
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Solution: \u03b1 = \u2212 \u2212 \u22c5 = \u2212 \u2212 \u22c5 = \u2212 M M M M M M M M mix mix mix mix mix ( ) ( ) n 1 2 1 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u2212 M dRT P PM dRT 1 1
Answer: B
\nOriginal PDF solution page
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Solution: n RT RT NO = \u00d7 = 0 4 250 100 . n RT RT O 2 0 8 100 80 = \u00d7 = . 2 NO + O 2 \u2192 2 NO 2 \u001f N 2 O 4 Initial moles 100 RT 80 RT 0 0 Final moles 0 30 RT 100 RT x \u2212 x 2 From question, 30 100 2 0 3 350 RT RT x x RT + \u2212 + = \u00d7 . \u2234 x RT = 50 Now, K P of second reaction = P P x RT x RT N O 2 4 2 2 2 1 2 100 0 3 0 3 350 NO = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u2212 . . = 3.5 atm
Answer: A
\nOriginal PDF solution page
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Solution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 1 3 0 Equilibrium moles 1 \u2013 x 3 \u2013 3 x 2 x From question, 2 4 2 x x a \u2212 =\n6.41 Chemical Equilibrium HINTS AND EXPLANATIONS \u21d2 x a a = + 2 1 Now, K x x x P x x x x P P = \u2212 \u2212 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2212 \u2212 \u22c5 \u2212 ( ) ( )( ) ( ) ( ) 2 1 3 3 4 2 4 4 2 27 1 2 3 2 2 2 4 2 or, K x x x P a a a a a P = \u22c5 \u2212 \u2212 \u22c5 = \u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u2212 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u2212 2 4 2 27 1 2 2 1 4 4 1 27 1 2 1 2 ( ) ( ) + + \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 a P 2 = \u22c5 \u2212 \u22c5 32 27 1 2 a a P ( ) \u2234 a a P ( ) 1 2 \u2212 \u03b1
Answer: B
\nOriginal PDF solution page
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Solution: K P P = \u22c5 \u2212 \u03b1 \u03b1 2 2 1 \u21d2 ( . ) ( ) . 0 3 1 1 0 3 0 1 1 2 2 2 2 \u00d7 \u2212 \u2212 = \u00d7 \u2212 \u03b1 \u03b1 \u21d2 a = 0.973
Answer: A
\nOriginal PDF solution page
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Solution: K P = P 2 Now, ln K K H R T T 2 1 1 2 1 1 = \u0394 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, ln ( ) P 2 2 3 2 7 10 3360 2 1 300 1 400 \u00d7 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u21d2 P 2 = 1.4 \u00d7 10 \u20132 atm
Answer: B
\nOriginal PDF solution page
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Solution: CO(g) + H 2 (g) \u001f CO 2 (g) + H 2 (g) Initial 1 5 0 1 Equilibrium 1 \u2013 x 5 \u2013 x x 1 \u2013 x Now, K x x x x eq = = \u22c5 + \u2212 \u22c5 \u2212 1 3 1 1 5 ( ) ( ) ( ) \u21d2 x = 1 2
Answer: D
\nOriginal PDF solution page
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Solution: NH 2 COONH 4 (s) \u001f N 2 + 3H 2 + CO + 1 2 2 O Equilibrium partial pressure 22 5 5 4 . = 3 22 5 5 12 \u00d7 = . 22 5 5 4 . = 22 2 5 5 2 \u00d7 = . Now, K p = \u00d7 \u00d7 \u00d7 = \u00d7 4 12 4 2 27 2 3 1 2 10 5 ( ) ( ) ( ) / .
Answer: A
\nOriginal PDF solution page
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Solution: NH 2 COONH 4 (s) \u001f 2NH 3 (g) + CO 2 (g) Equ. partial pressure 2 P 0 P 0 New Equ. partial pressure 3 P 0 P 0 Now, K P P P P P = \u22c5 = \u22c5 ( ) ( ) 2 3 0 2 0 0 2 \u21d2 P P = 4 9 0 Now, 3 3 31 27 0 0 P P P + =
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 H \u00b0 = \u0394 E \u00b0 + \u0394 n g \u22c5 RT = (+30) + (3 \u2013 2) \u00d7 2 1000 300 30 6 \u00d7 = + . K cal Now, \u0394 G \u00b0 \u2013 RT \u22c5 ln K eq = \u0394 H \u00b0 \u2013 T \u22c5 \u0394 S \u00b0 or, \u2013 2 \u00d7 300 \u00d7 ln K eq = 30.6 \u00d7 10 3 \u2013 300 \u00d7 100 \u21d2 ln K eq = \u2234 K eq = 1 e
Answer: B
\nOriginal PDF solution page
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Solution: trans \u001f Cis ; \u0394 G \u00b0 = 22.112 \u2013 30.426 = \u2013 8.314 KJ Now, \u0394 \u00b0 = \u2212 \u22c5 G RT Cis trans ln [ ] [ ] \u21d2 \u2013 8.314 \u00d7 10 3 = \u2013 8.314 \u00d7 300 \u00d7 ln [ ] [ ] Cis trans \u2234 [ ] [ ] Cis trans = 28 1
Answer: D
\nOriginal PDF solution page
Open page 326 in PDFExtracted text
Solution: CO( g ) + H 2 O( g ) \u001f CO 2 ( g ) + H 2 ( g ) Initial moles 2 5 0 2 Equilibrium moles 2 \u2013 x 5 \u2013 x x 2 + x Now, K x x x x eq = = \u22c5 + \u2212 \u2212 3 0 2 2 5 . ( ) ( )( ) \u21d2 x = 1.5 \u2234 [ ] . H M 2 2 2 1 75 = + = x
Answer: A
\nOriginal PDF solution page
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Solution: NH 4 HS(s) \u001f NH 3 (g) + H 2 S(g) X 2 bar X 2 bar \u2234 \u0394 G \u00b0 = \u2013 RT \u22c5 ln K P \u00b0 = \u2013 RT ln X X RT 2 2 2 2 , (ln ln ) \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u00d7 \u2212\n6.42 Chapter 6 HINTS AND EXPLANATIONS
Answer: C
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+ {
+ "title": "Chem Sec 2",
+ "originalName": "Section B - Multi Correct",
+ "questions": [
+ {
+ "question_id": "chemical-equilibrium-chem-sec-2-1-61",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": 1,
+ "subject": "chemistry",
+ "chapter": "chemical-equilibrium",
+ "chapterTitle": "Chemical Equilibrium",
+ "type": "mcqm",
+ "rawChapterType": "MSQ",
+ "originalNumber": 61,
+ "displayNumber": 1,
+ "image": "./assets/chapter-packs/chemical-equilibrium/Chemistry Section 1__--__61__--__1.png",
+ "solutionImage": null,
+ "question": {
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+ "identifier": "A",
+ "content": ""
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+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A",
+ "B",
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: A, B, C
\nOriginal PDF solution page
Open page 327 in PDFExtracted text
Solution: NH 4 HS(s) \u001f NH 3 (g) + H 2 S(g) Equ. partial pressure 0.2 atm 0.2 atm Second Equ. partial pressure 0.5 atm P atm Now, K P = 0.2 \u00d7 0.2 = 0.5 \u00d7 P \u21d2 P = 0.08
Answer: A, B
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Solution: N 2 O 5 (g) \u001f 2NO 2 (g) + 1 2 2 O ( ) g Initial partial pressure P 0 0 0 Equ. partial pressure P 0 (1 \u2013 a ) 2 a P 0 \u03b1 P 0 2 and \u03b1 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 = \u2212 M D D M D D 2 5 2 1 2 2 3 Total equilibrium pressure = P 0 (1 \u2013 a ) + 2 a P 0 + \u03b1 \u03b1 P P 0 0 2 1 3 2 = + \u239b \u239d \u239c \u239e \u23a0 \u239f
Answer: A, B
\nOriginal PDF solution page
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Solution: Le Chatelier\u2019s principle
Answer: A, B, D
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Solution: Vapour pressure of a particular liquid system depends only on temperature.
Answer: B, C
\nOriginal PDF solution page
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Solution: Cl 2 (g) \u001f 2Cl(g) T \u2191 P \u2193
Answer: B, C
\nOriginal PDF solution page
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Solution: Addition of insert gas at constant pressure shifts the equilibrium in the direction of increase in moles of gases.
Answer: A, B, D
\nOriginal PDF solution page
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Solution: Decrease in pressure favors the reaction is the direction of increase in moles of gas and hence, B should be monomer.
Answer: A, B
\nOriginal PDF solution page
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Solution: \u0394 \u00b0 > > f G : NO N O NO 2 2 5
Answer: A, D
\nOriginal PDF solution page
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Solution: At 300 K : \u0394 G \u00b0 = (\u201341) \u2013 300 \u00d7 (\u20130.04) = \u201329 KJ/mol Hence, the reaction is spontaneous in forward direction. At 1200 K : \u0394 G \u00b0 = (\u201333) \u2013 1200 \u00d7 (\u20130.03) = + 3 KJ/mol Hence, the reaction is spontaneous in backward direction.
Answer: B, C
\nOriginal PDF solution page
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Answer: A, B
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Solution: S A e H RT = \u22c5 \u2212\u0394 / \u21d2 ln s = ln A H RT \u2212 \u0394 Positive slope represents that \u0394 H = negative.
Answer: A, B, C
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Solution: K P P g 2 2 0 2 = = Cl atm ( ) . K P P P g 1 2 8 8 25 9 0 2 0 001 2 10 = \u22c5 = \u00d7 = \u00d7 \u2212 Cl H O(g) 2 atm ( ) . ( . ) P H O(g) 2 = Vapour pressure of ice.
Answer: A, C
\nOriginal PDF solution page
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Solution: PCl 5 (g) \u001f PCl 3 (g) + Cl 2 (g) Initial moles 5 0 0 Moles at equilibrium 5 \u2013 x x x From question, ( ) . . 5 4 4 8 112 0 0821 546 \u2212 + + + = \u00d7 \u00d7 x x x \u21d2 x = 3 \u2234 \u03b1 = = x 5 0 6 . and K x x x P = \u22c5 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = 5 4 8 12 1 8 . . atm
Answer: A, B, C, D
\nOriginal PDF solution page
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Solution: 4HCl(g) + O 2 (g) \u001f 2Cl 2 (g) + 2H 2 O(g) Initial partial pressure 1.0 atm 0.25 atm 0 0.4 atm On completion 0 0 0.5 atm 0.4 atm Equ. partial pressure 4 x atm x atm 0.5 atm 0.4 atm K x x P = \u00d7 = \u00d7 5 10 0 5 0 4 4 12 2 2 4 ( . ) ( . ) ( ) \u21d2 x = 5 \u00d7 10 \u20134
Answer: A, B
\nOriginal PDF solution page
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Solution: \u0394 \u00b0 = \u00d7 \u0394 \u00b0 \u2212 \u0394 \u00b0 = G G G f g f g 2 0 2 4 NO N O 2 ( ) ( ) \u21d2 K P \u00b0 = 1 Now, \u0394 = \u0394 \u00b0 + \u22c5 = + \u22c5 G G Q RT P P RT ln NO N O 2 4 0 2 2 ln = \u22c5 RT ln 10 10 2 = positive.\n6.43 Chemical Equilibrium HINTS AND EXPLANATIONS
Answer: B, C
\nOriginal PDF solution page
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Solution: AB 2 (g) + A( s ) \u001f 2 AB(g) Initial partial pressure 0.7 bar 0 Equ. partial pressure (0.7 \u2013 x ) bar 2 x bar Second equ. partial pressure y bar (0.4 \u2013 y ) bar From question, (0.7 \u2013 x ) + 2 x = 0.95 \u21d2 x = 0.25 \u2234 K x x P = \u2212 = = ( ) ( . ) ( . ) . 2 0 7 0 5 0 45 5 9 2 2 Now, 5 9 0 4 2 = \u2212 ( . ) y y \u21d2 y = 0.13 \u2234 At second equilibrium, the volume per cent of AB 2 0 13 0 4 100 32 5 = \u00d7 = . . . %
Answer: A, D
\nOriginal PDF solution page
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Solution: PCl 5 ( g ) \u001f PCl 3 ( g ) + Cl 2 ( g ) Initial moles 1 1 0 Equ. moles 1 \u2013 x 1 + x x \u2248 1 \u2248 1 = 0.004 \u2234 K C = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1 0 004 1 1 10 0 0004 . . M
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 \u00b0 = \u2212 \u22c5 \u00b0 G RT K P ln \u21d2 \u20131743 = \u2013 8.3 \u00d7 300 \u00d7 ln K P \u00b0 \u2234 K P \u00b0 = 2
Answer: A, C, D
\nOriginal PDF solution page
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Solution: K K K 1 2 3 1 1 0 24 = \u00d7 = . As \u0394 n g = 0, [ A ] + [ B ] + [ C ] = 1 M
Answer: C, D
\nOriginal PDF solution page
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Solution: Addition of water will shift the reaction in the direction of increase in mole of aq species.
Comprehension 1 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 328 in PDFExtracted text
Solution: For SrCl 2 \u22c5 2H 2 O(s), P H O 2 = (2.56 \u00d7 10 \u201310 ) 1/4 = 0.004 atm For Na 2 HPO 4 \u22c5 7H 2 O P H O 2 = (2.43 \u00d7 10 \u201313 ) 1/5 = 0.003 atm For Na 2 SO 4 (s), P H O 2 = (1.024 \u00d7 10 \u201327 ) 1/10 = 0.002 atm As P H O 2 is minimum for Na 2 SO 4 (s), it is the best dehydrating agent.
Comprehension 1 - subquestion 2
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 328 in PDFExtracted text
Solution: For Na 2 SO 4 (s), 10H 2 O(s) to be efflorescent, P H O 2 < 0.002 atm or 0 04 100 0 002 . . . . \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f < R H \u21d2 R . H . < 5 %
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 328 in PDFExtracted text
Solution: Na 2 HPO 4 \u22c5 7H 2 O(s) to be deliquescent, P H O 2 > 0.003 atm or, 0 04 100 0 003 . . . . \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f > R H \u21d2 R . H . > 7.5 % Comprehension II
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: CO( g ) + 2H 2 ( g ) \u001f CH 3 OH( g ) Initial moles 0.2 a (say) 0 Moles at equ. 0.2 \u2013 x a \u2013 2 x x = 0.1 = a \u2013 0.2 = 0.1 Total moles = 0.1 + ( a \u2013 0.2) + 0.1 = 7 5 2 463 0 0821 750 . . . \u00d7 \u00d7 \u21d2 a = 0.3 Now, K P = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 0 1 0 1 0 1 7 5 0 3 0 16 2 2 2 . . ( . ) . . . atm
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: K K RT C P n g = = \u00d7 = \u0394 \u2212 \u2212 ( ) . ( . ) 0 16 0 0821 750 607 2 2 M
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 328 in PDFExtracted text
Solution: P = + \u00d7 \u00d7 = ( . . ) . . . 0 2 0 3 0 0821 750 2 463 12 5 atm\n6.44 Chapter 6 HINTS AND EXPLANATIONS Comprehension III
Comprehension 3 - subquestion 1
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 329 in PDFExtracted text
Solution: C 2 H 5 OH + CH 3 COOH \u001f CH 3 COOC 2 H 5 + H 2 O Initial moles a a 0 0 Equilibrium moles a \u2013 x a \u2013 x x x From question, x a 2 0 333 1 3 = = . \u21d2 x a = 2 3 Now, K x x a x eq a x = \u22c5 \u2212 = \u22c5 \u2212 ( ) ( ) 4
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 329 in PDFExtracted text
Solution: C 2 H 5 OH + CH 3 COOH \u001f CH 3 COOC 2 H 5 + H 2 O Initial moles a 3 2 3 a 0 0 Equilibrium moles a x 3 \u2212 2 3 a x x x Now, K x x a x a x eq = = \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 4 3 2 3 \u21d2 x = 0.2833 a \u2234 Fraction of alcohol reacted = x a / . 3 0 85 =
Comprehension 3 - subquestion 3
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 329 in PDFExtracted text
Solution: Solution of 0.7 = x a \u00d7 = 100 66 67 . % Comprehension IV
Comprehension 4 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 329 in PDFExtracted text
Solution: 2HI( g ) \u001f H 2 ( g ) + I 2 ( g ) Initial moles 1(say) 0 0 Equilibrium moles 1 \u2013 0.2222 = 0.7778 0.1111 0.1111 \u2234 K eq = \u00d7 = \u2248 0 1111 0 1111 0 7778 1 49 0 02 2 . . ( . ) .
Comprehension 4 - subquestion 2
",
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+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: In the presence of I 2 ( g ), the extent of dissociation of HI will decrease.
Comprehension 4 - subquestion 3
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: Addition of He( g ) will not affect thequilibrium. Comprehension V
Comprehension 5 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 329 in PDFExtracted text
Solution: NH 4 HS ( s ) \u001f NH 3 ( g ) + H 2 S ( g ) Initial partial pressure P mm 0 Equilibrium partial pressure ( P + x )mm x mm From question, P + x = 625 and ( P + x ) + x = 725 \u2234 x = 100 and P = 525
Comprehension 5 - subquestion 2
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: K P = ( P + x ) \u22c5 x = 625 \u00d7 100 mm 2 \u2234 \u2032 K P (required) = 1 1 6 10 5 K P = \u00d7 \u2212 . mm \u20132 .
Comprehension 5 - subquestion 3
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 329 in PDFExtracted text
Solution: P P K P NH H S mm 3 2 250 = = =
Comprehension 6 - subquestion 1
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 329 in PDFExtracted text
Solution: Minimum mass of NH 4 HS ( s ) needed. = \u00d7 \u00d7 \u00d7 250 760 5 0 0 0821 300 51 . . gm\n6.45 Chemical Equilibrium HINTS AND EXPLANATIONS Comprehension VI
Comprehension 6 - subquestion 2
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 330 in PDFExtracted text
Solution: K A A e e e eq f b H RT = \u22c5 = \u2212 \u2212 \u00d7 \u00d7 = \u2212\u0394 / ( . ) . 24 942 10 8 314 300 3 10
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: K K K e b f eq = = \u2212 1 10 \u0394 H = Ea f \u2013 Ea b = Ea f \u2013 3 2 Ea f \u21d2 Ea f = 2 \u22c5 (\u2013 \u0394 H) Now, K A e e e f f Ea RT f = \u22c5 = \u00d7 = \u2212 \u2212 \u00d7 \u00d7 \u00d7 \u2212 / . . 1 2 24 942 10 8 314 300 20 3 and K b = e \u201330 Comprehension VII
Comprehension 7 - subquestion 1
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: 2SO 3 \u001f 2SO 2 + O 2 Equilibrium moles 1 \u2013 a a \u03b1 2 Now. K K P P = \u22c5 \u2212 \u22c5 + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u2032 \u03b1 \u03b1 \u03b1 \u03b1 2 2 2 1 1 2 ( ) \u21d2 \u03b1 = 2 3
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: 2NH 3 \u001f N 2 + 3H 2 Equilibrium moles 1 \u2212 \u03b1 \u03b1 2 3 2 \u03b1 = 1 3 = 1 3 = 1 Now, K P = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f = 1 3 1 1 3 50 5 3 2700 3 2 2 2 atm
Comprehension 7 - subquestion 3
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: n initial \u00d7 (17 + 28 + 2 + 20) = 134 \u21d2 n initial = 2 2NH 3 \u001f N 2 + 3H 2 Initial moles 2 2 2 Moles at equilibrium 2 \u2013 2 x 2 + x 2 + 3 x = 1.0 = \u00d7 134 0 5224 28 . = 3.5 \u2234 x = 0.5 Now, K P P = = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 2700 2 5 3 5 1 0 9 3 2 2 . ( . ) ( . ) \u21d2 P = \u00d7 \u00d7 2700 81 2 5 3 5 3 . ( . ) atm Comprehension VIII N 2 + 3H 2 \u001f 2NH 3 ; K P 1 Initial partial pressure 9 P 13 P 0 Equilibrium partial pressure 9 P \u2013 x \u2013 y 13 P \u2013 3 x \u2013 2 y 2 x N 2 + 2H 2 \u001f N 2 H 4 ; K P 2 Initial partial pressure 9 P 13 P 0 Equilibrium partial pressure 9 P \u2013 y \u2013 x 13 P \u2013 2 y \u2013 3 x y From question, P x P NH 3 2 0 = = \u21d2 x P = 0 2 P P x y P H 2 13 3 2 2 0 = \u2212 \u2212 = and P total = (9 P \u2013 x \u2013 y ) + (13 P \u2013 3 x \u2013 2 y ) + 2 x + y = 7 P 0 \u2234 y P = 3 2 0 and P P = 0 2 K P P P P P P P P 1 3 2 2 2 3 0 2 0 0 3 0 2 5 2 2 1 20 = \u22c5 = \u00d7 = NH N H ( ) \u2234 K P (required) = 20 0 2 P
Comprehension 8 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: K P P P P P P P P 2 2 4 2 2 2 0 0 0 2 0 2 3 2 5 2 2 3 20 = \u22c5 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = N H N H ( )\n6.46 Chapter 6 HINTS AND EXPLANATIONS
Comprehension 8 - subquestion 2
",
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+ "identifier": "A",
+ "content": ""
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+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
"
+ }
+ },
+ {
+ "question_id": "chemical-equilibrium-chem-sec-3-24-93",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": 1,
+ "subject": "chemistry",
+ "chapter": "chemical-equilibrium",
+ "chapterTitle": "Chemical Equilibrium",
+ "type": "mcq",
+ "rawChapterType": "MCQ",
+ "originalNumber": 93,
+ "displayNumber": 24,
+ "image": "./assets/chapter-packs/chemical-equilibrium/Chemistry Section 1__--__93__--__1.png",
+ "solutionImage": null,
+ "question": {
+ "content": "Comprehension 8 - subquestion 3
",
+ "options": [
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
"
+ }
+ }
+ ]
+ },
+ {
+ "title": "Chem Sec 4",
+ "originalName": "Section D - Assertion Reason",
+ "questions": [
+ {
+ "question_id": "chemical-equilibrium-chem-sec-4-1-94",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": 1,
+ "subject": "chemistry",
+ "chapter": "chemical-equilibrium",
+ "chapterTitle": "Chemical Equilibrium",
+ "type": "mcq",
+ "rawChapterType": "MCQ",
+ "originalNumber": 94,
+ "displayNumber": 1,
+ "image": "./assets/chapter-packs/chemical-equilibrium/Chemistry Section 1__--__94__--__1.png",
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+ "identifier": "C",
+ "content": ""
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+ {
+ "identifier": "D",
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+ "D"
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+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: Direction of shifting of equilibrium will depend on relative values of a and b .
Answer: A
\nOriginal PDF solution page
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Solution: Equilibrium opposes the changes.
Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: K K RT P C n g = \u22c5 \u0394 ( )
Answer: B
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: Exothermic direction is favoured on lowering temperature.
Answer: A
\nOriginal PDF solution page
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Solution: NaCl( s ) \u001f Na + ( aq ) + Cl \u2013 ( aq )
Answer: C
\nOriginal PDF solution page
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Solution: On decreasing the volume, moles of A( g ) as well as B( s ) will increase.
Answer: A \u2192 S; B \u2192 Q, R; C \u2192 Q; D \u2192 P
\nOriginal PDF solution page
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Answer: A \u2192 S; B \u2192 R; C \u2192 P; D \u2192 Q
\nOriginal PDF solution page
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Solution: K K RT P C n g = \u22c5 \u0394 ( )
Answer: A \u2192 P, R; B \u2192 S; C \u2192 Q; D \u2192 S
\nOriginal PDF solution page
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Solution: Le Chatelier\u2019s principle
Answer: A \u2192 P, Q, S; B \u2192 P, Q; C \u2192 R
\nOriginal PDF solution page
Open page 331 in PDFExtracted text
Solution: Le Chatelier\u2019s principle
Answer: A \u2192 P, Q, R, S; B \u2192 Q, R, S; C \u2192 P, Q, R, S, T; D \u2192 Q, R
\nOriginal PDF solution page
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Solution: Le Chatelier\u2019s principle
Answer: A \u2192 S, T; B \u2192 R; C \u2192 Q; D \u2192 P
\nOriginal PDF solution page
Open page 331 in PDFExtracted text
Solution: Le Chatelier\u2019s principle
Answer: A \u2192 P, S, T; B \u2192 Q, R, S; C \u2192 S; D \u2192 Q, R, S
\nOriginal PDF solution page
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Solution: Le Chatelier\u2019s principle
Answer: A \u2192 P, S, T; B \u2192 Q, R; C \u2192 Q, R; D \u2192 P, S
\nOriginal PDF solution page
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Solution: Le Chatelier\u2019s principle
Answer: A \u2192 P, R; B \u2192 Q, R; C \u2192 Q, S
\nOriginal PDF solution page
Open page 331 in PDFExtracted text
Solution: P K P CO atm 2 2 463 = = . \u2234 n CO 2 at equilibrium = \u00d7 \u00d7 = 2 463 15 0 0821 900 0 5 . . . (A) % of CaCO 3 decomposed = \u00d7 = 0 5 1 0 100 50 . . % ( ) Eqn (B) % of CaCO 3 decomposed = \u00d7 = 0 5 0 5 100 100 . . % ( ) Eqn (C) % of CaCO 3 decomposed = 100% (non ) -Eqn
Answer: A \u2192 Q, R, S; B \u2192 P, R, S; C \u2192 P, R, S; D \u2192 Q, R, S
\nOriginal PDF solution page
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Solution: Le Chatelier\u2019s principle
Answer: 2
\nOriginal PDF solution page
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Solution: 2H 2 S(g) \u001f 2H 2 (g) + S(g); K e = 10 \u20136 Initial moles 0.1 0 0 Equilibrium moles 0.1 \u2013 x x x 2 \u001a 0 1 . Now, K x x C = = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 10 2 0 1 1 0 4 6 2 2 ( . ) . \u21d2 x = 2 \u00d7 10 \u20133 \u2234 Percentage dissociation = \u00d7 \u00d7 = \u2212 2 10 0 1 100 2 3 . %
Answer: 4
\nOriginal PDF solution page
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Solution: V V CF CO ml 4 2 500 300 200 = \u2212 = = \u2234 V COF ml 2 500 2 200 100 = \u2212 \u00d7 = Hence, P P CF CO atm 4 2 200 500 10 4 = = \u00d7 = P COF atm 2 100 500 10 2 = \u00d7 = K p = \u00d7 = 4 4 2 4 2
Answer: 7
\nOriginal PDF solution page
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Solution: Initial: n PCl 5 62 55 208 5 0 3 = = . . . and n Cl 2 4 48 22 4 0 2 = = . . . PCl 5 \u001f PCl 3 + Cl 2 Initial moles 0.3 0 0.2 Equilibrium moles 0.3 \u2013 x x 0.2 + x\n6.47 Chemical Equilibrium HINTS AND EXPLANATIONS Now, K p = x x x p x x x x RT V \u22c5 + \u2212 \u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u2212 \u00d7 ( . ) ( . ) . ( . ) ( . ) 0 2 0 3 0 5 0 2 0 3 or, 8 0 2 0 3 0 0821 546 4 48 = + \u2212 \u00d7 \u00d7 x x x ( . ) . . . \u21d2 x = 0.2 \u2234 Final pressure = + \u00d7 \u00d7 = ( . ) . . 0 5 0 0821 546 4 48 7 x atm
Answer: 1
\nOriginal PDF solution page
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Solution: NaOH is used to neutralize acetic acid. From the given data, half of the acid taken is neutralize. CH 3 COOH + C 2 H 5 OH \u001f CH 3 COOC 2 H 5 + H 2 O Initial moles a a 0 0 Equilibrium moles a a \u2212 2 a a \u2212 2 a 2 a 2 \u2234 K a a a a eq = \u00d7 \u00d7 = 2 2 2 2 1
Answer: 2
\nOriginal PDF solution page
Open page 332 in PDFExtracted text
Solution: 2HI \u001f H 2 + I 2 Equilibrium moles 1 \u2013 0.8 0.4 0.4 = 0.2 \u2234 K eq = \u00d7 = 0 4 0 4 0 2 4 2 . . ( . ) Now, H 2 + I 2 \u001f 2HI Initial moles 2 2 0 Equilibrium moles 2 \u2013 x 2 \u2013 x 2 x K x x x eq = = \u2212 \u2212 1 4 2 2 2 2 ( ) ( )( ) \u21d2 x = 0.4 Now, n eq of I 2 = n eq of Na 2 S 2 O 3 or, (2 \u2013 x ) \u00d7 2 = V \u00d7 (1.6 \u00d7 1) \u21d2 V = 2 L
Answer: 5
\nOriginal PDF solution page
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Solution: Cl 2 CHCOOH + C 5 H 10 \u001f Cl 2 CHCOOC 5 H 11 I: Initial moles 1 4 0 Equilibrium moles 1 \u2013 x 4 \u2013 x x = 0.5 II: Initial moles 1 a 0 Equilibrium moles 1 \u2013 y a \u2013 y y = 0.6 Now, K a eq = \u00d7 \u00d7 = \u00d7 \u2212 \u00d7 0 5 0 5 3 5 0 7 0 6 0 4 0 6 0 72 . . . . . . ( . ) . \u21d2 a = 5\n6.48 Chapter 6 HINTS AND EXPLANATIONS
Answer: 5
\nOriginal PDF solution page
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Solution: n CO2 at equilibrium = 0.05 \u2234 Minimum mass of CaCO 3 needed = 0.05 \u00d7 100 = 5 gm
Answer: 5
\nOriginal PDF solution page
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Solution: Ag + (aq) + Fe 2+ (aq) \u001f Fe 3+ (aq) + Ag(s) Initial moles 500 0 9 1000 0 45 \u00d7 = . . 500 1 0 1000 0 50 \u00d7 = . . 0 0 Equilibrium moles 0.45 \u2013 x 0.50 \u2013 x x x Now, n eq Fe 2+ = n eq MnO 4 \u2212 or, ( . ) . 0 50 1000 30 1 25 0 06 1000 5 \u2212 \u00d7 \u00d7 = \u00d7 \u00d7 x \u21d2 x = 0.25 \u2234 K x x x eq M = \u2212 \u2212 = \u2212 ( . )( . ) 0 45 0 5 5 1
Answer: 1
\nOriginal PDF solution page
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Solution: Sb2S3(s) + 3H2(g) \u001f 2Sb(s) + 3H2S(g) Initial moles 0.01 0.01 0 0 Equ. moles 0.01 \u2013 x 0.01 \u2013 3 x 2 x 3 1 19 238 x = . = 5 \u00d7 10 \u20133 = 5 \u00d7 10 \u20133 Now, K c = \u00d7 \u00d7 = \u2212 \u2212 ( ) ( ) 5 10 5 10 1 3 3 3 3
Answer: 4
\nOriginal PDF solution page
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Solution: H 2 O + D 2 O \u001f 2HDO Initial moles 28 28 0 Equ. moles 28 \u2013 14 = 14 28 \u2013 14 = 14 2 \u00d7 14 = 28 K C = \u00d7 = ( ) 28 14 14 4 2
Answer: 6
\nOriginal PDF solution page
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Solution: 3A 2 (g) \u001f A 6 (g), K p 1 1 6 2 = \u2212 . atm Initial partial pressure 2 P 0 0 Equilibrium partial pressure 2 P 0 \u2013 3 a \u2013 b a A 2 (g) + C (g) \u001f A 2 C (g), K x p 2 1 = \u2212 atm Initial partial pressure 2 P 0 P 0 0 Equilibrium partial pressure 2 P 0 \u2013 b \u2013 3 a P 0 \u2013 b b From question, a = 0.2, P P P A A A 6 2 2 3 3 1 6 0 2 1 6 = \u21d2 = . . . \u21d2 P P a b A 2 0 5 2 3 0 = = \u2212 \u2212 . and (2 P 0 \u2013 3 a \u2013 b ) + a + ( P 0 \u2013 b ) + b = 1.4 \u21d2 P 0 = 0.7 and b = 0.3 Now, K b P a b P b p 2 2 3 0 3 0 5 4 1 5 0 0 1 = \u2212 \u2212 \u2212 = \u00d7 = \u2212 ( )( ) . . . . atm
Answer: 4
\nOriginal PDF solution page
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Solution: Initial partial pressure of IBr (g) = \u00d7 \u00d7 = 8 28 207 0 0821 500 0 1642 10 . . . 2IBr (g) \u001f I 2 (g) + Br 2 (g) Initial partial pressure 10 0 0 Equilibrium partial pressure 10 \u2013 2 x x x = 4 \u2234 K p = \u00d7 = 4 4 2 4 2 ( )
Answer: 4
\nOriginal PDF solution page
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Solution: H 2 (g) + I 2 (g) \u001f 2HI (g) I: Initial moles 1 3 0 Equilibrium moles 1 2 \u2212 x 3 2 \u2212 x x II: Initial moles 3 3 0 Equilibrium moles 3 \u2013 x 3 \u2013 x 2 x\n6.49 Chemical Equilibrium HINTS AND EXPLANATIONS Now, K x x x x x x eq = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 ( ) ( ) ( )( ) 2 2 1 2 3 2 2 3 3 \u21d2 x = 3 2 \u2234 K eq = 4
Answer: 1
\nOriginal PDF solution page
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Solution: N 2 O 5 (g) \u001f N 2 O 3 (g) + O 2 (g); K C 1 2 5 = . M Initial moles 4 0 0 Equilibrium moles 4 \u2013 x x \u2013 y x + y N 2 O 3 (g) \u001f N 2 O(g) + O 2 (g); K C 2 x 0 0 Equilibrium moles x \u2013 y y x + y From question, [ ] . O 2 2 2 5 = + = x y \u21d2 x + y = 5 And K x y x y y C 1 2 5 5 4 1 2 5 2 5 1 1 2 = = \u2212 \u00d7 \u2212 \u00d7 = \u2212 \u00d7 \u2212 \u00d7 . ( ) ( ) ( ) \u21d2 y = 2 \u2234 [N O] 2 2 1 = = y M
Answer: 1
\nOriginal PDF solution page
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Solution: In left chamber, P H e = 2 atm \u2234 P P NH H 3 2 atm = = \u2212 = 3 4 2 2 1 \u2234 K p = \u00d7 = 1 1 1 2 atm Four-digit Integer Type
Answer: 0480
\nOriginal PDF solution page
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Solution: P-xyloquinone + M.W \u001f P-xylohydroquinone + M.B. Initial conc. 0.012 M 0 0.24 M 10 \u20133 M Equ. con. 0.012 + 4 \u00d7 10 \u20135 4 \u00d7 10 \u20135 M 0.24 \u2013 4 \u00d7 10 \u20135 10 4 100 10 3 3 \u2212 \u2212 \u2212 \u00d7 \u2248 0.012 \u2248 0.24 M 0.96 \u00d7 10 \u20133 m \u2234 K eq = 0 24 0 96 10 0 012 4 10 480 3 5 . . . \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212
Answer: 0015
\nOriginal PDF solution page
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Solution: 6HCHO \u001f C 6 H 12 0 6 ; K eq = 6.4 \u00d7 10 19 Initial conc. 0 1 M Equ. con. 6 x 1 \u2013 x = 1 M Now, K eq = 6.4 \u00d7 10 19 = 1 6 [ ] HCHO \u21d2 [HCHO] = 5 \u00d7 10 \u20134 M = 5 \u00d7 10 \u20134 \u00d7 30 g/L = 15 mg/L
Answer: 0600
\nOriginal PDF solution page
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Solution: PCl 5 \u001f PCl 3 + Cl 2 Initial equ. moles 2 2 2 Moles on adding Cl 2 2 2 2 + x Final Equ. moles 2 + y 2 \u2013 y 2 + x \u2013 y From question, (2 + y) + (2 \u2013 y ) + 2 + ( x \u2013 y ) = 2 \u00d7 6 or, x \u2013 y = 6 and K c = 2 2 2 1 2 2 2 1 2 \u00d7 \u00d7 = \u2212 \u00d7 + \u2212 + \u00d7 V y x y y V ( ) ( ) ( ) Or, 4 = ( ) ( ) 8 8 4 \u2212 \u00d7 \u2212 x x \u21d2 x = 20 3
Answer: 1784
\nOriginal PDF solution page
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Solution: K \u00b0 = eq 1 \u21d2 \u2206 G \u00b0 = 0 \u21d2 T = \u0394 \u0394 H S \u00b0 \u00b0 = \u00d7 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 223 10 223 33 1520 10 3 3 = 1784 K
Answer: 0180
\nOriginal PDF solution page
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Solution: Graphite \u001f Diamond; \u2206 G\u00b0 = (3.0 \u2013 0) kJ/mol; P 1 = 1 bar \u2206 G = 0 P 2 = P Now, \u2206 ( \u2206 G ) = \u2206 V \u2219 \u2206 P or, ( \u2206 G \u2013 \u2206 G \u00b0) = ( V Dia \u2013 V Gra ) ( P 2 \u2013 P 1 ) or, (0 \u2013 3.0 \u00d7 10 3 ) = 12 3 6 12 2 4 10 10 6 5 2 . . \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 P \u2234 P 2 = 1.8 \u00d7 10 9 Pa = 1.8 \u00d7 10 4 bar
Answer: 0532
\nOriginal PDF solution page
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Solution: K P P P \u00b0 = = CO CO 2 10 400 6 Now, \u2206 G \u00b0 = \u20135320 \u2013 5.6 T = \u2013RT ln K P \u00b0\n6.50 Chapter 6 HINTS AND EXPLANATIONS = \u20132 \u00d7 T \u00d7 ln 10 400 6 \u2234 T = 532 k
Answer: 0032
\nOriginal PDF solution page
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Solution: 4HNO 3 (g) \u001f 4NO 2 (g) + 2H 2 O (g) + O 2 (g) Initial partial pressure P 0 0 0 0 Equ. partial pressure P 0 \u2013 4 x 4 x 2 x x From question, P 0 \u2013 4 x = 2 atm and ( P 0 \u2013 4 x ) + 4 x + 2 x + x = 30 atm \u2234 P 0 = 18 atm and x = 4 atm Now, K x x x P x p o = \u00d7 \u00d7 \u2212 = ( ) ( ) ( ) 4 2 4 2 4 2 4 20 3 atm and K K RT c p n g = ) = \u00d7 = \u0394 ( ( . ) 2 0 08 400 32 20 3 M 3
Answer: 0032
\nOriginal PDF solution page
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Solution: Initial: P NOcl = P 0 bar and P N 2 = (1 \u2013 P 0 ) bar 2NOCl \u001f 2NO + Cl 2 Initial partial pressure P 0 0 0 Eqn. partial pressure P 0 \u2013 2 x 2 x x = 1.2\u20131.0 = 0.2 Par. pre. on adding Cl 2 P 0 \u2013 2 x 2 x x + (8.3 \u2013 1.2) New Equ.partial pre. P 0 \u2013 2 x + 2 y 2 x \u2013 2 y 7.1 + x \u2013 y From question, y = 8.3 \u2013 8.2 = 0.1 Now, K x x P x x y x y P x y p o = \u00d7 \u2212 = \u2212 \u00d7 + \u2212 \u2212 + ( ) ( ) ( ) ( . ) ( ) 2 2 2 2 7 1 2 2 2 2 2 0 2 or, 0 4 0 2 0 4 0 2 7 2 0 2 0 5 2 0 2 2 0 2 0 . . ( . ) . . ( . ) . \u00d7 \u2212 = \u00d7 \u2212 \u21d2 = P P P and K p = 3.2
Answer: 0170
\nOriginal PDF solution page
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Solution: Initial equilibrium: A + 2B \u001f C Initial partial pressure P 0 2 P 0 O Equ. partial pressure P 0 \u2013 x 2 P 0 \u2013 2 x x From question, ( P 0 \u2013 x ) + (2 P 0 \u2013 2 x ) + x + 3 P 0 = 5 6 6 0 \u00d7 P or, x = P 0 2 Second equilibrium: A + 2B \u001f C Initial partial pressure 2 P 0 4 P 0 0 Equ. partial pressure 2 P 0 \u2013 y 4 P 0 \u2013 2 y y \u2013 2 z 2C + D \u001f 2F y 6 P 0 0 Equ. partial pressure y \u2013 2 z 6 P 0 \u2013 z 2 z From question, 2 P 0 \u2013 y = y \u2013 2 z Now for the first reaction, K x P x P x y z P y P y p o = \u2212 \u2212 = \u2212 \u2212 \u2212 ( )( ) ( )( ) 2 2 2 2 4 2 0 2 0 0 2 or, 1 1 4 2 3 2 2 0 2 0 2 0 0 P P y y P z P = \u2212 \u21d2 = = ( ) and Total equilibrium pressure: First equilibrium = 5 P 0 Second equilibrium = (2 P 0 \u2013 y ) + (4 P 0 \u2013 2y) + ( y \u2013 2 z ) + (6 P 0 \u2013 z ) + 2 z = 3.5 P 0
Answer: 0018
\nOriginal PDF solution page
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Solution: A 2 + B 2 \u001f 2AB Initial moles 0.2 0.2 0 Equ. moles 0.2 \u2013 x 0.2 \u2013 x 2 x = 0.3 = 0.05 = 0.05 K eq = K 1 = ( . ) . . 0 3 0 05 0 05 36 2 \u00d7 = After adding C 2 : A 2 + B 2 \u001f 2AB Initial moles 0.2 0.2 0 Equ. moles 0.2 \u2013 y \u2013 z 0.2 \u2013 y 2 y = 0.24 = 0.08 \u2013 z = 0.08 A 2 + C 2 \u001f 2AC Initial moles 0.2 0.1 0 Equ. moles 0.2 \u2013 z \u2013 y 0.1 \u2013 z 2 z Now, K z z 1 2 36 0 24 0 08 0 08 0 06 = = \u2212 \u00d7 \u21d2 = ( . ) ( . ) . . \u2234 K eq = K 2 = ( . ) . . 0 12 0 02 0 04 18 2 \u00d7 =
Answer: 2400
\nOriginal PDF solution page
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Solution: \u0394 G = \u0394 G \u00b0 + RT. ln Q = \u2013RT. ln K K f b + RT [Product] [Reactants] .ln\n6.51 Chemical Equilibrium HINTS AND EXPLANATIONS = RT. ln K K b f [Product] [Reactants] = RT. ln r r b f = 2 \u00d7 300 \u00d7 ln 1 4 e = \u20132400 cal
Answer: 0250
\nOriginal PDF solution page
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Solution: A + B \u001f C; K 1 = 4 \u00d7 10 10 Initial moles 5 5 0 Equ. moles 5 \u2013 ( x + y ) 5 \u2013 x x A + D \u001f C; K 2 = 10 10 Initial moles 5 5 0 Equ. moles 5 \u2013 ( x + y ) 5 \u2013 y y As K 1 and K 2 are very large, ( x + y ) = 5 (1) and K K x x y y x y 1 2 4 5 5 2 = = \u2212 \u00d7 \u2212 \u21d2 = (2) From (1) and (2), x = 10 3 \u2234 Moles of B at equilibrium = 5 \u2013 x = 5 3
Answer: 0800
\nOriginal PDF solution page
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Solution: Br 2 (l) + Cl 2 (g) \u001f 2Br Cl (g); K p = 1 atm Initial moles x 10 0 Equ. moles \u2248 0 10 \u2013 x 2 x Br 2 (l) \u001f Br 2 (g); K p = 0.25 atm Initial moles y 0 Equ. moles \u2248 0 y From question: y y \u00d7 \u00d7 = \u21d2 = 0 082 300 164 0 25 5 3 . . and ( ) . . 10 2 0 082 300 164 2 00 10 3 \u2212 + \u00d7 \u00d7 = \u21d2 = x x x \u2234 Minimum mass of Br 2 (l) = ( x + y ) \u00d7 160 gm = 800 gm
Answer: 0016
\nOriginal PDF solution page
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Solution: 2SO 3 \u001f 2SO 2 + O 2 Initial moles 1 (say) 0 0 Equ. moles 1 \u2013 0.4 = 0.6 0.4 0.2 \u2234 M av = 1 80 1 2 \u00d7 . Now, d = PM RT p p \u21d2 = \u00d7 \u00d7 \u21d2 = 16 80 1 2 0 0821 920 0 0821 216 . . . atm \u2234 K p = \u00d7 \u00d7 = ( . ) . ( . ) . 0 4 0 2 0 6 216 1 2 16 2 2 atm
Answer: 0008
\nOriginal PDF solution page
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Solution: A 2 \u001f 2A; K 1 = x atm Initial partial pressure 1 atm 0 Equ. partial pressure 1 \u2013 ( x + z ) 2 x B 2 \u001f 2B; K 2 = y atm Initial partial pressure 1 atm 0 Equ.partial pressure 1 \u2013 ( y + z ) 2 y A 2 + B 2 \u001f 2AB; K 3 = 2 Initial partial pressure 1 1 0 Equ. partial pressure 1 \u2013 ( x + z ) 1 \u2013 ( y + z ) 2 z = 0.5 (1) From question, [1 \u2013 ( x + z )] + 2 x + [1 \u2013 ( y + z )] + 2 y + 2 z = 2.75 \u2234 x + y = 0.75 (2) Now, K 3 = ( . ) ( . )( . ) . . 0 5 0 75 0 75 2 0 25 0 50 2 \u2212 \u2212 = \u21d2 = x y x or y = 0.50 or 0.25 \u2234 K K y y z x x z y x x y 2 1 2 2 2 2 2 1 2 1 2 0 75 2 0 75 = \u2212 + \u2212 + = \u00d7 \u2212 \u00d7 \u2212 ( ) ( ) ( ) ( ) ( ) ( . ) ( ) ( . ) = = 1 8 1 or 8
![\"Chemical](\"./assets/chapter-packs/chemical-kinetics/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Negative sign is for reactants and positive for products.
Answer: B
\nOriginal PDF solution page
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Solution: r d dt K K rxn = \u2212 \u22c5 = \u2212 1 2 2 1 2 2 2 2 4 [ ] [ ] [ ] NO NO N O \u2234 Rate of disappearance of NO 2 is given by, \u2212 = \u2212 d dt K K [ ] [ ] [ ] NO NO N O 2 1 2 2 2 2 4 2 2
Answer: D
\nOriginal PDF solution page
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Solution: P n RT V A A = \u21d2 dP dt RT V dn dt A A = \u22c5 or, ( ) ( ) \u2212 \u22c5 = \u2212 \u22c5 K P RT K C A n A n 1 2 \u2234 K K RT P C K RT RT RT A A n n n 2 1 1 1 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 = \u2212 ( ) ( )
Answer: D
\nOriginal PDF solution page
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Solution: \u2212 \u22c5 = \u2212 1 2 2 2 1 2 2 d dt K K [ ] [ ] [ ][ ] HI HI H I
Answer: C
\nOriginal PDF solution page
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Solution: Order = 1 \u2190\u23af with respect to A \u21d2 a = 1 Order = 2 \u2190\u23af with respect to B \u21d2 b = 2 Hence, reaction is A + 2B P. r d A dt d B dt = \u2212 = \u2212 [ ] [ ] 1 2
Answer: C
\nOriginal PDF solution page
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Solution: A + 2B C + D t = 0 0.6 atm 0.8 atm t = t 0.6 \u2013 x 0.8 \u2013 2 x = 0.3 atm = 0.2 atm \u21d2 x = 0.3 \u2234 r r K K t 0 2 2 0 3 0 2 0 6 0 8 1 32 = \u00d7 \u00d7 \u00d7 \u00d7 = . ( . ) . ( . )\n11.45 Chemical Kinetics HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: For no change in temperature, \u0394 H net = 0 and hence, for 3 moles of B reacted, 4 moles of Q should form.
Answer: A
\nOriginal PDF solution page
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Solution: 2.82 = (2) x \u21d2 x = 3 2 9 = (3) y \u21d2 y = 2 \u2234 overall order = x + y = 7 2
Answer: D
\nOriginal PDF solution page
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Solution: r 1 = 0.0068/65 = 1.046 \u00d7 10 \u20134 gm/min r 2 = 0.0031/120 = 2.583 \u00d7 10 \u20135 gm/min r 3 = 0.0032/60 = 5.333 \u00d7 10 \u20135 gm/min From (1) and (2) : order with respect to K 2 C 2 O 4 = 2 From (1) and (3) : order with respect to HgCl 2 = 1
Answer: D
\nOriginal PDF solution page
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Solution: From (2) and (3) : order with respect to I \u2013 = 1 From (1) and (3) : order with respect to ClO \u2013 = 1 From (3) and (4) : order with respect to OH \u2013 = 1
Answer: B
\nOriginal PDF solution page
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Solution: (1 + K 2 \u22c5 C A ) \u2248 1
Answer: A
\nOriginal PDF solution page
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Solution: For steady state, + = d R dt [ ] 0 or, K 1 [ A ] \u2013 K 2 [ R ][ B ] \u2013 K 3 [ R ][ C ] = 0 \u2234 [ ] [ ] [ ] [ ] R K A K B K C = + 1 2 3 Now, dx dt K R C K K A C K B K C = = + 3 3 1 2 3 [ ][ ] [ ][ ] [ ] [ ]
Answer: C
\nOriginal PDF solution page
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Solution: r K = 3 3 [ ][ ] O O For 1st step, K K 1 2 2 3 = [ ][ ] [ ] O O O \u2234 r = K 3 [O][O 3 ] = K K K 3 1 3 2 2 2 [ ] [ ] O O
Answer: D
\nOriginal PDF solution page
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Solution: r d X dt d Y dt = \u2212 = + [ ] [ ] and rate decreases with time.
Answer: C
\nOriginal PDF solution page
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Solution: A 2B t = 0 0.1 M 0 t = 1 min 0.1 \u2013 x 2 x For zero order reaction: [ A 0 ] \u2013 [ A ] = Kt or, 0.1 \u2013 (0.1 \u2013 x ) = 0.01 \u00d7 1 \u21d2 x = 0.01 \u2234 [ B ] = 2 x = 0.02 M
Answer: B
\nOriginal PDF solution page
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Solution: 2NH 3 N 2 + 3H 2 r r r r rxn = = = = = NH N H atm/s Constant 3 2 2 2 1 3 0 1 . Hence, after 10 seconds: P NH atm 3 3 2 0 1 10 1 = \u2212 \u00d7 \u00d7= . P N atm 2 0 1 10 1 = \u00d7 = . P H atm 2 3 0 1 10 3 = \u00d7 \u00d7 = . \u2234 P total = 1 + 1 + 3 = 5 atm
Answer: B
\nOriginal PDF solution page
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Solution: For \u2212 = d A dt K A n [ ] [ ] and n \u2260 1 [ A ] 1 \u2013 n = [ A 0 ] 1 \u2013 n \u2013 K (1 \u2013 n ) \u22c5 t For given graph, 1 \u2013 n = \u20133 \u21d2 n = 4 and \u2013 K (1 \u2013 n ) = tan 45\u00b0 \u21d2 K (4 \u2013 1) = 1 \u2234 K = \u2212 \u2212 1 3 3 1 M min Now, r d A dt K A rxn = \u2212 \u22c5 = \u22c5 \u22c5 1 3 1 3 4 [ ] [ ] = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 1 3 1 3 0 2 16 9 10 4 4 1 ( . ) min M
Answer: C
\nOriginal PDF solution page
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Solution: 1 0 25 2 8 0 1 2 M M hr t t = \u00d7 \u23af \u2192 \u23af\u23af\u23af / . . \u21d2 t 1/2 = 4.0 hr 0.6M M hr t t = \u23af \u2192 \u23af\u23af 1 2 4 0 0 3 / . .
Answer: C
\nOriginal PDF solution page
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Solution: The successive t 1/2 are double of previous one and hence, order =
Answer: C
\nOriginal PDF solution page
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Solution: 20. r = K [ A ] n 10 = K (0.8) n (1) 0.625 = K (0.2) n (2) \u2234 n = 2
Answer: A
\nOriginal PDF solution page
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Solution: C 2 H 6 C 2 H 4 + H 2 t = 0 3 bar 0 0 t = ? (3 \u2013 x ) bar x bar x bar From question: (3 \u2013 x ) + x + x = 5 \u21d2 x = 2 From the unit of rate constant, the order of reaction is 2, hence, t K P P x = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00b0 1 1 1 1 0 0015 1 3 1 3 1 10 2 6 2 6 5 C H C H . = 4.44 \u00d7 10 \u20133 hr = 16 seconds
Answer: B
\nOriginal PDF solution page
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Solution: Let the reaction be first-order. K 1 1 8 100 20 0 201 = \u22c5 = ln . K 2 1 18 100 10 0 128 = \u22c5 = ln .\n11.46 Chapter 11 HINTS AND EXPLANATIONS As K 1 \u2260 K 2 , the reaction is not first-order. Let the reaction be second-order. K 1 1 8 1 0 2 1 1 0 5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = . . K 2 1 18 1 0 1 1 1 0 5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = . . As K 1 = K 2 , order = 2
Answer: D
\nOriginal PDF solution page
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Solution: Here, t 1/2 is independent from sugar concentration and hence, the order with respect to sugar is
Answer: D
\nOriginal PDF solution page
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Solution: Now, r = K [Sugar][H + ] n = K \u2032 \u22c5 [Sugar] t K K n 1 2 2 2 / ln ln [ ] = \u2032 = + H 500 2 10 5 = \u22c5 \u2212 ln ( ) K n and 50 2 10 6 = \u22c5 \u2212 ln ( ) K n \u2234 n = \u20131
Answer: C
\nOriginal PDF solution page
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Solution: r = K [ester] [ H + ] = K \u2032 \u22c5 [ester] \u2234 t K K 1 2 2 2 0 693 0 1 0 01 693 / ln ln [ ] . . . = \u2032 = = \u00d7 = + H hr
Answer: C
\nOriginal PDF solution page
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Solution: For n th order reaction ( n \u2260 1) Kt A A n n n = \u2212 \u2212 \u2212 \u2212 [ ] [ ] 0 1 1 1 For n = 0.5, Kt A A = \u2212 [ ] [ ] / / 0 1 2 1 2 1 2 Now, t T A K A K 100 0 1 2 1 2 0 1 2 2 0 2 % / / / ([ ] ) [ ] = = \u2212 = and t t A A K A 50 1 2 0 1 2 0 1 2 0 1 2 2 2 2 1 1 2 % / / / / [ ] [ ] [ ] = = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239f K \u2234 T t 1 2 1 1 1 2 1 0 3 / . = \u2212 =
Answer: A
\nOriginal PDF solution page
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Solution: For A : t t A A B A = \u22c5 = \u22c5 1 2 0 0 2 10 2 8 / log log [ ] [ ] log log [ ] [ ] For B : t t B B B A = \u22c5 = \u22c5 1 2 0 0 2 20 2 / log log [ ] [ ] log log [ ] [ ] From question, 10 2 8 20 2 0 0 log log [ ] [ ] log log [ ] [ ] \u22c5 = \u22c5 B A B A \u2234 [ ] [ ] B A 0 8 = \u21d2 t B A = \u22c5 = 20 2 60 0 log log [ ] [ ] min Alternate method: A B B B B B B : [ ] [ ] [ ] [ ] [ ] [ ] 8 4 2 2 4 0 10 0 10 0 10 0 10 0 10 0 10 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u23af \u2192 \u23af [ ] B 0 8 B B B B B : [ ] [ ] [ ] [ ] 0 20 0 20 0 20 0 2 4 8 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af
Answer: A
\nOriginal PDF solution page
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Solution: 0 1 0 025 2 40 1 2 . . / min M M t t = \u23af \u2192 \u23af\u23af \u21d2 t 1/2 = 20 min Now, r K A t A = = = \u00d7 [ ] ln [ ] . min . / 2 0 693 20 0 01 1 2 M = 3.465 \u00d7 10 \u20134 M min \u20131
Answer: A
\nOriginal PDF solution page
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Solution: + = \u2212 = d B dt d A dt K A [ ] [ ] [ ] / 1 3 or, \u2212 = \u22c5 \u222b \u222b d A B K dt A A t [ ] [ ] / [ ] [ ]/ / 1 3 2 0 0 0 1 2 t A K 1 2 0 2 3 2 3 5 3 3 2 1 2 / / / / [ ] ( ) = \u2212 \u22c5
Answer: C
\nOriginal PDF solution page
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Solution: N 2 O 5 2NO 2 + 1 2 O 2 t = 0 a mole 0 t = t ( a \u2013 x ) mole x V t 2 mole \u03b1 t = \u221e \u001f 0 a V 2 mole \u03b1 \u221e K t t a a x t V V V t = \u22c5 = \u22c5 \u2212 = \u22c5 \u2212 \u221e \u221e 1 1 1 2 5 0 2 5 ln [ ] [ ] ln ln N O N O Now, 1 20 9 6 9 6 4 8 1 40 9 6 9 6 \u22c5 \u2212 = \u22c5 \u2212 ln . . . ln . . V t \u21d2 V t = 7.2 ml
Answer: A
\nOriginal PDF solution page
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Solution: For zero order reaction, t P 1 2 3 / \u03b1 NH \u00b0 \u2234 315 70 150 1 2 t / = \u21d2 t 1/2 = 675 sec
Answer: A
\nOriginal PDF solution page
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Solution: A n B t = 0 P 0 0 t = t P 0 \u2013 x n.x Now, ( P 0 \u2013 x ) = P 0 \u22c5 e \u2013 Kt \u21d2 x = P 0 (1 \u2013 e \u2013 Kt )\n11.47 Chemical Kinetics HINTS AND EXPLANATIONS Now, P total = ( P 0 \u2013 x ) + nx = P 0 \u22c5 e \u2013 Kt + n \u22c5 P 0 (1 \u2013 e \u2013 Kt ) = P 0 [ n + (1 \u2013 n ) e \u2013 Kt ]
Answer: B
\nOriginal PDF solution page
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Solution: t K n K n n n n n 1 2 1 1 1 1 2 1 1 2 1 / ( ( ) ( ) [ ] ( ) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 C ) C C 0 0 0 and t K n K n n n n n 3 4 1 1 1 2 1 4 1 1 2 1 / ( ) ( ( ) ( ) [ ] ( ) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 C ) C C 0 0 0 \u2234 t t n n n 3 4 1 2 2 1 1 1 1 2 1 2 1 2 / / ( ) ( ) ( ) = \u2212 \u2212 = + \u2212 \u2212 \u2212
Answer: D
\nOriginal PDF solution page
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Solution: r r A A n 2 1 2 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f [ ] [ ] \u21d2 2 = (4) n \u21d2 n = 1 2 Now, t 1/2 a [ A 0 ] 1 \u2013 n \u21d2 t 1/2 a [ A 0 ] 1/2 100 50 25 16 16 2 t t = = \u23af \u2192 \u23af\u23af\u23af \u23af \u2192 \u23af\u23af\u23af\u23af min / min \u2234 Time for 75 % reaction = 16 16 2 27 3 + = . min
Answer: D
\nOriginal PDF solution page
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Solution: Kt a a x x a = \u2212 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f ln ln 1 1 or, 2.5 \u00d7 10 \u20135 \u00d7 (100 \u00d7 60) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f ln 1 1 x a \u21d2 x a = 0 138 . \u2234 Percentage decomposition = \u00d7 = x a 100 13 8 . %
Answer: A
\nOriginal PDF solution page
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Solution: ( )( ) / / t t 1 2 1 1 2 2 = \u21d2 0 693 1 1 2 0 . [ ] K K A = \u2234 [ ] . . . . . A K K 0 1 2 2 0 693 6 93 10 0 693 0 2 0 5 = = \u00d7 \u00d7 = \u2212 M
Answer: A
\nOriginal PDF solution page
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Solution: t t t t t t 1 2 1 2 1 1 2 2 1 2 1 1 2 2 100 75 2 100 25 = \u22c5 \u22c5 = ( ) log log ( ) log log ( ) ( / / / / ) ) log log 2 4 3 4 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 = 3 2 0 6 0 48 0 6 3 10 . . .
Answer: B
\nOriginal PDF solution page
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Solution: r = K [ A ] n 1 100 0 02 60 0 02 \u00d7 = . ( . ) K n (1) 1 100 0 04 15 0 04 \u00d7 = . ( . ) K n (2) \u2234 n = 3
Answer: B
\nOriginal PDF solution page
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Solution: t T A A = \u22c5 gen ln ln [ ] [ ] 2 0 \u21d2 60 75 2 0 = \u22c5 ln ln [ ] [ ] A A \u2234 [ ] [ ] . A A e 0 0 56 =
Answer: A
\nOriginal PDF solution page
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Solution: ( ) ( ) [ ] [ ] / / t t A A n 1 2 1 1 2 2 0 1 0 2 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u21d2 37 82 18 95 0 05 0 10 1 . . . . = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 n \u21d2 n = 2 Now, ( ) . . . / t 1 2 1 37 82 0 15 0 05 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u21d2 t 1/2 = 12.6 hr
Answer: C
\nOriginal PDF solution page
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Solution: Now, T N a 0 1000 2 \u22c5 = \u00d7 (1) and T N a x x x t \u22c5 = \u2212 \u00d7 + \u00d7 + \u00d7 1000 2 2 1 ( ) (2) \u2234 a a x T T T t \u2212 = \u2212 0 0 3 2
Answer: A
\nOriginal PDF solution page
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Solution: B n + B ( n + 4)+ t = 0 a mole 0 t = 10 min ( a \u2013 x ) mol x mol Now, 25 1000 2 \u00d7 = \u00d7 N a (1) and, 32 5 1000 2 5 . ( ) \u00d7 = \u2212 \u00d7 + \u00d7 N a x x (2) Now, K t a a x = \u22c5 \u2212 = \u22c5 \u2212 = \u2212 1 1 10 12 5 12 5 2 5 0 02 1 ln min ln . . . . min CH (Br) COOH CH (Br) COOH a mole 0 x mole 0 x mole ( a \u2013 x ) mole t = 0 t = t CHCOOH C Br COOH + H Br \u2192\n11.48 Chapter 11 HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: A 2B + C t = 0 a mol 0 0 t = 10 sec ( a \u2013 x ) mol 2 x mol x mol Now, r r P P M M A B A B B A = \u22c5 \u21d2 1 2 2 4 16 = \u2212 \u22c5 a x x \u21d2 x a = 3 Now, K t a a x a a a = \u22c5 \u2212 = \u22c5 \u2212 = \u2212 1 1 10 2 3 0 04 1 ln sec ln . sec
Answer: A
\nOriginal PDF solution page
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Solution: r = K [ A ] 2 [ B ] = K \u2032 \u22c5 [ A ] 2 as [ B 0 ] >> [ A 0 ] \u2234 t K A K B A 1 2 0 0 0 1 1 1 0 5 0 002 2 0 500 / [ ] [ ][ ] . . . min = \u2032 \u22c5 = = \u00d7 \u00d7 =
Answer: D
\nOriginal PDF solution page
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Solution: r = K [ester][H + ] \u2234 r r HA HX HA H = = + 1 100 1 0 [ ] . \u21d2 [H + ] HA = 0.01 M Now, Ka A HA ( ) [ ][ ] [ ] . . ( . ) HA H = = \u00d7 \u2212 \u2248 + \u2212 \u2212 0 01 0 01 1 0 01 10 4
Answer: A
\nOriginal PDF solution page
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Solution: As [ A 0 ] = [ B 0 ] and the stoichiometric coefficients of both A and B are 1, at any time [ A ] = [ B ]. Hence, r = K [ A ] 1/2 [ B ] 1/2 = K [ A ]. Required time = 2 2 0 693 2 31 10 600 1 2 3 \u00d7 = \u00d7 \u00d7 = \u2212 t / . . sec
Answer: C
\nOriginal PDF solution page
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Solution: \u2212 = + dC dt C C \u03b1 \u03b2 1 \u21d2 \u2212 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 = \u22c5 \u222b \u222b 1 0 2 1 2 C dC dt t Co Co \u03b2 \u03b1 / / \u2234 t C 1 2 0 1 2 2 / ln = + \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u03b1 \u03b2
Answer: C
\nOriginal PDF solution page
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Solution: r = K \u2032 [CH 3 COOH][C 2 H 5 OH] = K \u2032 \u22c5 [CH 3 COOH] 2 \u2234 t K 1 2 0 1 / [ = \u2032 \u22c5 CH COOH] 3 \u21d2 50 1 10 0 2 3 = \u00d7 \u00d7 \u2212 ( ) . K \u2234 K = 100 M \u20132 min \u20131
Answer: C
\nOriginal PDF solution page
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Solution: r = K [ A ] x [ B ] y For case-I : r = K [ A ] x [ B ] y = K \u2032 \u22c5 [ B ] y where K \u2032 = K [ A 0 ] x In equal time interval, the concentrations of B are in G.P. and hence, y = 1 and \u2032 = \u22c5 = \u2212 K 1 10 0 01 0 008 0 02 1 ln . . . min For case-II: r = K [ A ] x [ B ] y = K \u2033 [ A ] x where K \u2033 = K [ B 0 ] y In equal time interval, the concentration of A are in G.P. and hence, x = 1 and \u2032\u2032 = \u22c5 = \u2212 K 1 10 0 02 0 018 0 01 1 ln . . . min Now, r = K [ A ][ B ] \u2234 K K A K B = \u2032 \u2032\u2032 = = \u2212 \u2212 [ ] , [ ] . . . . . min 0 0 1 1 0 02 2 0 0 01 1 0 0 01 or or M
Answer: B
\nOriginal PDF solution page
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Solution: K K C C K C app = \u22c5 + \u22c5 = + 1 1 1 1 \u03b1 \u03b1 lim C K K \u2192\u221e = app 1 \u03b1 From question, K C C K 1 1 1 90 100 \u22c5 + \u22c5 = \u00d7 \u03b1 \u03b1 or, C C 1 9 10 90 100 1 9 10 5 5 + \u00d7 = \u00d7 \u00d7 \u21d2 C = 10 \u20135 M
Answer: A
\nOriginal PDF solution page
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Solution: A 2B + C t = 0 1 0 0 t = 12 hr 1 \u2013 x 2 x x t = 24 hr 1 \u2013 y 2 y y V.P. of solution, P = X 2 \u22c5 P o or, 20 180 18 180 18 1 2 24 = + + \u00d7 / ( ) x \u21d2 x = 0.5 \u2234 t = 12 hr = t 1/2 Now, t = 24 hr = 2 \u00d7 t 1/2 \u21d2 y = 0.75 Now, V.P. of solution, P X P y = \u22c5 \u00b0 = + + \u00d7 2 10 10 1 2 24 ( ) = 19.2 mm Hg
Answer: B
\nOriginal PDF solution page
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Solution: [ ] [ ] . . B C K K = = \u00d7 \u00d7 = \u2212 \u2212 1 2 4 5 1 26 10 3 15 10 4 1 \u2234 Percentage of B = \u00d7 = 4 5 100 80%\n11.49 Chemical Kinetics HINTS AND EXPLANATIONS
Answer: A
\nOriginal PDF solution page
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Solution: A R t = t a \u2013 x x \u2234 r = K ( a \u2013 x ) \u22c5 x For maximum rate, dr dx = 0 \u21d2 x a = 2 \u21d2 C A = C R
Answer: A
\nOriginal PDF solution page
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Solution: K dt by y dy t \u22c5 = + \u2212 \u22c5 \u222b \u222b 0 0 2 1 2 1 / / Co Co \u21d2 t K b b 1 2 1 1 2 2 / ( ) ln = + \u22c5 \u22c5 \u2212 \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 Co Co
Answer: A
\nOriginal PDF solution page
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Solution: t t B A 50 94 %, %, = or, 1 100 50 1 100 6 2 1 K K \u22c5 = \u22c5 ln ln \u21d2 K K 1 2 4 067 1 = .
Answer: A
\nOriginal PDF solution page
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Solution: Percentage product by S N 2 mechanism = \u00d7 \u00d7 + \u00d7 \u00d7 \u2212 \u2212 \u2212 ( . )[ ]( . ) ( . )[ ]( . ) . [ ] 4 8 10 0 01 4 8 10 0 01 2 4 10 100 5 5 6 RX RX RX = = 16 67 . %
Answer: B
\nOriginal PDF solution page
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Solution: [H 2 ] : [O 2 ] : [OH] : [H 2 O] : [O] = K 1 : K 1 : 2 K 2 : K 3 : K 3 = 0.60 : 0.60 : 2 \u00d7 0.30 : 0.10 : 0.10 = 6 : 6 : 6 : 1 : 1
Answer: A
\nOriginal PDF solution page
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Solution: [ A ] + [ B ] + [ C ] = [ A 0 ] when [ A ] = [ B ] = [ C ], [ A ] = [ ] [ ] A A e Kt 0 0 3 = \u22c5 \u2212 or, 1 3 3 3 = \u2212 + \u22c5 e t (ln ln ) \u21d2 t = 0.5 min
Answer: D
\nOriginal PDF solution page
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Solution: As K 1 = K 2 = K (Say), t K max . min = = = 1 1 0 02 50 and [ ] [ ] . max B A e e = = 0 0 2 M
Answer: B
\nOriginal PDF solution page
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Solution: After long time, r A = r B \u21d2 K 1 [ A ] = K 2 [ B ] \u2234 [ ] [ ] A B K K = = 2 1 40
Answer: A
\nOriginal PDF solution page
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Solution: [ ] [ ] [ ] ( ) [ ] ( ( ) ( ) C A K A K K e A e K K K e K K t K K t = + \u2212 \u22c5 = + \u2212 + \u2212 + \u22c5 2 0 1 2 0 2 1 2 1 1 2 1 2 ( ( ) ) K K t 1 2 1 + \u22c5 \u2212 = \u2212 = \u2212 \u22c5 \u00d7 \u00d7 \u00d7 \u2212 9 10 1 9 10 1 1 1 10 10 1 25 10 3600 1 5 K K e e K t ( ) [ ] . = 0.5112
Answer: C
\nOriginal PDF solution page
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Solution: At steady state, K 1 [ A ] = K 2 [ B ] \u2234 K K A B 2 1 4 3 1 2 5 10 0 2 0 01 5 10 = = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 [ ] [ ] . . . min
Answer: B
\nOriginal PDF solution page
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Solution: Reaction may be considered as A C K 1 \u23af \u2192 \u23af \u2234 [ C ] = [ A 0 ] ( ) 1 1 \u2212 \u2212 e K t
Answer: A
\nOriginal PDF solution page
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Solution: A B K 1 2 \u23af \u2192 \u23af A C K 2 \u23af \u2192 \u23af t = 0 1 atm 0 1 atm 0 t = 10 min (1 \u2013 x \u2013 y ) 2 x 1 \u2013 x \u2013 y y t = \u221e (1 \u2013 a \u2013 b ) 2 a (1 \u2013 a \u2013 b ) b \u2248 0 \u2248 0 From question, a + b = 1 and 2 a + b = 1.5 \u2234 a = b = 0.5 Now, P P K K a b x y B C = = = 2 2 2 1 2 \u21d2 K K x y 1 2 1 = = 0 Now, P x y x y 10 1 2 1 4 min ( ) . = \u2212 \u2212 + + = \u21d2 x = y = 0.4 \u2234 P x y A = \u2212 \u2212 = 1 0 2 . atm at t = 10 min Now, K 1 + K 2 = 1 1 10 1 0 2 0 16 1 t P P A A \u22c5 \u00b0 = \u22c5 = \u2212 ln ln . . min \u2234 K 1 = K 2 = 0.08 min \u20131
Answer: A
\nOriginal PDF solution page
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Solution: r z u N av max * = = \u22c5 \u22c5 11 2 2 1 2 \u03c0\u03c3 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 1 2 4 10 2 10 2 10 8 2 4 1 19 3 2 \u03c0 ( ( ) ( ) cm) cm s cm = 2.842 \u00d7 10 28 cm \u20133 s \u20131 = 4.74 \u00d7 10 7 mol l \u20131 s \u20131
Answer: B
\nOriginal PDF solution page
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Solution: d K dT E RT a (ln ) = 2 or, 0 2 2 + + = \u03b2 \u03b3 T T E RT a \u21d2 E a = ( b T + \u03b3 ) R
Answer: A
\nOriginal PDF solution page
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Solution: K K B C 1 2 40 60 2 3 = = = [ ] [ ] Now, E K E K E K K a a a ( ) overall = \u22c5 + \u22c5 + 1 1 2 2 1 2 = 32 kcal/mol\n11.50 Chapter 11 HINTS AND EXPLANATIONS
Answer: C
\nOriginal PDF solution page
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Solution: r r uncat cat = \u00d7 1 2 \u21d2 K K uncat cat = \u00d7 1 2 or, A e A e E RT E RT T a a \u22c5 = \u00d7 \u22c5 \u2212 \u2212 \u00d7 ( ) ( ) / / uncat cat 0.5 1 2 or, ln . ( ) ( ) 2 20 0 5 \u2212 = \u2212 \u2212 E RT E RT a a uncat uncat \u2234 E a (uncat) = 38.58 kcal/mol
Answer: B
\nOriginal PDF solution page
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Solution: For A B; K 1 = 8 min \u20131 at T = 300 K \u2032 = K 1 ? at T = ? ln \u2032 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f K R T 1 8 20 1 300 1 KJ (1) For A C; K 2 = 2 min \u20131 at T = 300 K \u2032 = K 2 ? at T = ? ln . \u2032 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f K R T 2 2 28 314 1 300 1 KJ (2) From (1) and (2), ln / / . . \u2032 \u2032 = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f K K T 2 1 3 2 8 8 314 10 8 314 1 300 1 or, ln 1 2 8 2 1 300 1 10 3 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 T \u21d2 T = 379.75 K
Answer: C
\nOriginal PDF solution page
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Solution: Given: K K 1 310 1 300 2 ( ) ( ) = , K 1 310 2 30 ( ) ln min = K K 1 310 1 310 2 ( ) ( ) = and E E a a 2 1 1 2 = For reaction 1: ln ( ) ( ) K K E R a 1 310 1 300 1 1 300 1 310 \u23a1 \u23a3 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f (1) For reaction 2: ln ( ) ( ) K K E R a 2 310 2 300 2 1 300 1 310 \u23a1 \u23a3 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f (2) From (1) \u00f7 (2) : ln ln ( ) ( ) 2 2 2 310 2 300 K K \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = or, K K 2 310 2 300 2 ( ) ( ) \u23a1 \u23a3 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 = \u21d2 K 2 (300) = K 2 310 2 2 2 30 2 ( ) = \u00d7 ln = 0.0327 min \u20131
Answer: B
\nOriginal PDF solution page
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Solution: At 27\u00b0C, K 1 1 1 21 6 100 25 2 10 8 = \u22c5 = \u2212 . ln ln . min Now, ln . . K K E R T T a 2 1 1 2 3 1 1 9 6 10 2 1 300 1 320 1 0 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2234 K K e 2 1 2 7 = = . \u21d2 K 2 2 7 2 10 8 2 4 = \u00d7 = . ln . ln \u21d2 ( t 1/2 ) 2 = 4 min \u2234 Percentage decomposition in 8.0 min = 75 %
Answer: C
\nOriginal PDF solution page
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Solution: K A e A e A e A E Rt RT RT a = \u22c5 = \u22c5 = \u2212 \u2212 / / . \u001f 0 37
Answer: A
\nOriginal PDF solution page
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Solution: ln 2 1 280 1 290 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E R a (1) and ln x E R a = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 290 1 300 (2) From (2) \u00f7 (1), ln ln x 2 280 300 = \u21d2 x = 1.91
Answer: B
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Answer: A, B, C, D
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Answer: B, C
\nOriginal PDF solution page
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Answer: A, C, D
\nOriginal PDF solution page
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Solution: r K A n = \u22c5 [ ] \u21d2 n r K A = ln( / ) ln [ ] Now, r r A A n 2 1 2 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f [ ] [ ] \u21d2 n r r A A = \u2212 \u2212 ln ln ln[ ] ln[ ] 2 1 2 1\n11.51 Chemical Kinetics HINTS AND EXPLANATIONS And, t 1/2 a [ A 0 ] 1\u2013 n \u21d2 ( ) ( ) [ ] [ ] / / t t A A n 1 2 2 1 2 1 0 2 0 1 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2234 n A A t t = \u2212 \u2212 \u2212 1 0 2 0 1 1 2 2 1 2 1 ln[ ] ln[ ] ln( )ln( ) / /
Answer: A, D
\nOriginal PDF solution page
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Solution: [ A ] = [ A 0 ] (1 \u2013 a ) = [ A 0 ] \u22c5 e \u2013 Kt \u21d2 \u03b1 = 1 \u2013 e \u2013 Kt
Answer: A, C
\nOriginal PDF solution page
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Solution: (a) \u2212 = \u22c5 d A dt K A n [ ] [ ] f A A A d A A = \u2212 = \u2212 [ ] [ ] [ ] [ ] [ ] 1 2 1 From question, f d A A = \u2212 [ ] [ ] \u2234 f A t K A n [ ] [ ] = \u21d2 f t K A n = \u22c5 \u2212 [ ] 1 or, log log ( ) log[ ] f t K n A \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u2212 \u22c5 1 (b) [ ] [ ] A A n Kt n n 0 1 1 1 \u2212 \u2212 \u2212 \u2212 = \u21d2 [ A ] 1 \u2013 n = [ A 0 ] 1 \u2013 n + ( n \u2013 1) \u22c5 Kt (c) t t A A A A n n n n 3 4 1 2 0 1 0 1 0 1 0 1 4 2 1 2 / / [ ] [ ] [ ] [ ] ( = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u2212 \u2212 \u2212 2 2 1 1 1 1 2 1 2 ) n n n \u2212 \u2212 \u2212 \u2212 = +
Answer: B, C, D
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Solution: t 1/2 = C \u22c5 (C 0 ) 1 \u2013 n \u21d2 ln t 1/2 = ln C + (1 \u2013 n ) \u22c5 ln C 0
Answer: B
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Solution: K \u2032 = K \u22c5 [H + ] On doubling [H + ], K \u2032 will double but K will remain unchanged.
Answer: D
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Answer: A, D
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Solution: (a) For steady state, 6 93 10 80 0 693 100 6 3 . . [ ] \u00d7 = \u00d7 \u2212 SO \u2234 [SO 3 ] = 1.25 \u00d7 10 \u20135 M (b) n eq SO 3 = n eq NaOH \u21d2 1.25 \u00d7 10 \u20135 \u00d7 10 3 \u00d7 2 = V NaOH \u00d7 1 \u2234 V NaOH = 2.5 \u00d7 10 \u20132 L = 25 ml (c) Mole of SO 3 needed = 980 10 98 10 3 4 \u00d7 = \u2234 Air needed = \u00d7 = \u00d7 \u2212 10 1 25 10 8 10 4 5 8 . L (d) 1000 days = 10 t 1/2 \u2234 [ ] . . SO M 3 5 10 8 1 25 10 2 1 25 10 = \u00d7 \u2248 \u00d7 \u2212 \u2212
Answer: C, D
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Solution: 3A( g ) 2B( g ) + 2C( s ) t = 0 6 atm 0 \u2013 t = 20 min (6 \u2013 x ) atm 2 3 x atm 0.05 atm t = \u221e \u2248 0 4 atm 0.05 atm But from question, P \u221e = 4.05 atm and hence, (4.05 \u2013 4) = 0.05 atm is the vapour pressure of C( s ). Now, P x x 20 6 2 3 0 05 5 05 = \u2212+ + = ( ) . . \u21d2 x = 2 \u2234 t = 20 min = t 1/2
Answer: A, B
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Solution: \u2212 = \u22c5 d dt K \u03b8 \u03b8 \u21d2 Kt = ln \u03b8 \u03b8 0 (a) t K = \u22c5 = \u22c5 = 1 1 0 04 596 298 17 5 0 ln . ln . sec \u03b8 \u03b8 (b) t = \u22c5 = 1 0 04 1192 298 35 . ln sec
Answer: A, C
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Solution: (a) A + B 2C t = 0 2 a a t = t 2 a \u2013 x a \u2013 x As [A] \u2260 [B] throughout, the overall reaction is not fi rst-order. (b) r = K [A] \u20131 [B] 2 = K \u2032 \u22c5 [B] 2 \u21d2 t K B 1 2 0 1 / [ ] = \u2032 (c) r = K [A] \u20131 [B] 2 = K \u2033 [A] \u20131 (d) As [A] = [B] = stoichiometric ratio, then the mole ratio will remain constant throughout.
Answer: B, D
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Solution: A P K 1 \u23af \u2192 \u23af ; t K 1 2 1 0 693 / . = B Q K 2 \u23af \u2192 \u23af ; t K B K 1 2 2 0 2 1 1 / [ ] = = From question, 0 693 1 1 2 . K K = \u21d2 K 2 > K 1
Answer: A, B, D
\nOriginal PDF solution page
Open page 578 in PDFExtracted text
Solution: A 4 4A t = 0 a M 0 t = 30 min ( a \u2013 x ) M 4 x M As a \u2013 x = 4 x \u21d2 x a = 5 \u2234 Percentage reaction at t = 30 min = \u00d7 = x a 100 20% Now, 30 2 1 2 = \u22c5 \u2212 t a a x / log log \u21d2 t 1/2 = 90 min\n11.52 Chapter 11 HINTS AND EXPLANATIONS
Answer: A, C, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: (a) \u0394 r H = \u2211 \u0394 f H Products \u2013 \u2211 \u0394 f H Reactants = 2 \u00d7 (\u20131263) \u2013 [(\u20132238) + (\u2013285)] = \u20133 KJ/mol (b) Can not confirm because in aqueous medium, there is no combustion. (d) Concentration in G.P. in equal time interval.
Answer: B, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: n = 1 \u21d2 t 100% = 1 0 0 K A \u22c5 = ln [ ] Infinite n \u2260 1 \u21d2 t 100 % = [ ] ( ) ( ) [ ] ( ) A K n A K n n n n n 0 1 1 0 1 0 1 1 \u2212 \u2212 \u2212 \u2212 \u2212 = \u2212 if < 1 = Infi nite if n > 1
Answer: A, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: [ ] [ ] B C K K = = 1 2 1 2 \u21d2 [ C ] > [ B ] Hence, after long time, the solution will be dextrorotatory. Now, K = K 1 + K 2 = 6.93 \u00d7 10 \u20132 + 13.86 \u00d7 10 \u20132 = 3 \u00d7 6.93 \u00d7 10 \u20132 min \u20131 \u2234 t K 1 2 2 2 0 693 3 6 93 10 10 3 / ln . . min = = \u00d7 \u00d7 = \u2212 A B A C t = 0 2M 0 2M 0 t = t 2 \u2013 ( x + y )M x M 2 \u2013 ( x + y )M y M From question, x + y = 1.5 and x y = 1 2 \u2234 x = 0.5, y = 1.0 Hence, total rotation = 0.5 \u00d7 60\u00b0 + 0.5 \u00d7 (\u201372\u00b0) + 1.0 \u00d7 42\u00b0 = 36\u00b0
Answer: A, B, C
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: [ B ] : [ C ] : [ D ] = 1 \u00d7 3 K : 2 \u00d7 2 K : 3 \u00d7 K = 3 : 4 : 3 A B A 2C A 3D t = 0 1M 0 1M 0 1M 0 t = t 1 \u2013 ( x + y + z ) x M 1 \u2013 ( x + y + z ) 2 y M 1 \u2013 ( x + y + z ) 3 z M t = \u221e 1 \u2013 ( a + b + c ) a M 1 \u2013 ( a + b + c ) 2 b M 1 \u2013 ( a + b + c ) 3 c M As a : 2 b : 3 c = 3 : 4 : 3 and a + b + c = 1 [ C ] = 2 b = 0.67 M As [ A 0 ] = 1 M, [ B ] \u2260 1M
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: For S N 1 path : r 1 = (3 \u00d7 10 \u20134 s \u20131 ) [RX] For S N 2 path : r 2 = (5 \u00d7 10 \u20134 M \u20131 s \u20131 ) [RX] [ ] \u001f\u001f Nu (a) [ ] \u001f\u001f Nu = 0.1 M, then r 1 > r 2 (b) [ ] \u001f\u001f Nu = 1.0 M, then r 1 < r 2 (c) [ ] \u001f\u001f Nu = 0.6 M, then r 1 = r 2 (d) [ ] \u001f\u001f Nu = 0.4 M, then r r 1 2 2 3 = \u2234 Percentage product by S N 1 = 2 2 3 100 40 + \u00d7 = %
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: \u2212 = + d A dt d B dt [ ] [ ] always
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: As mole is not changing, C A + C B + C C = C A 0 Now, C C C C C C K K K B A A B B C 0 1 1 2 \u2212 = + = +
Answer: A, C, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Answer: B, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Answer: A, C
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: Increase in temperature will result in greater increase in the rate of reaction A \u2192 B than B \u2192 C.
Answer: B
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Answer: C
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Answer: C, D
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: K 1 = K 2 \u21d2 \u2212 + = \u2212 + 14000 5 20000 10 RT RT \u21d2 T K = 1200 8 314 . Now, P P e e A B K t K t 2 3 1 2 1 1 1 1 = \u00d7 \u00d7 = \u2212 \u2212 Now, initial pressure P 0 1 1 0 0821 1200 8 314 100 0 237 = + \u00d7 \u00d7 = ( ) . . . atm As number of moles will increase on reaction, the total pressure can never be less than 0.2 atm Now, P P K K A B = = 2 3 2 3 1 2
Answer: A, B, C
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: 1 1 25 10 6 3 2 K dK dT d K dT T E RT a \u22c5 = = \u00d7 = (ln ) . \u2234 E R T a = \u00d7 = \u00d7 \u00d7 = 1 25 10 1 25 10 2 250 10 6 6 4 . . cal/mol
Answer: A, B, C
\nOriginal PDF solution page
Open page 579 in PDFExtracted text
Solution: \u0394 = \u2212 H E E a a f b \u21d2 \u2212 = \u2212 2 8 E a f \u21d2 E a f = 6 kcal/mol Now, the fraction of molecules crossing energy barrier = \u2212 e E RT a / and K e H Rt eq = \u2212\u0394 /\n11.53 Chemical Kinetics HINTS AND EXPLANATIONS
Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: The overall reaction is first-order.
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: K K K 1 2 3 2 4 1 = = \u21d2 2 K 1 = K 2 = 4 K 3
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: 2N 2 O 5 4NO 2 + O 2 2 \u00d7 108 gm 4 \u00d7 46 gm 32 gm 108 gm 92 gm 16 gm Comprehension II
Comprehension 1 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: CO(g) + Cl 2 (g) COCl 2 (g) r d dt K COCl 2 COCl COCl][Cl = + = \u22c5 [ ] [ ] 2 5 2 (1) Now, for steady state of COCl, + = d dt [ ] COCl 0 or K 3 [Cl][CO] \u2013 K 4 [COCl] \u2013 K 5 [COCl][Cl 2 ] = 0 \u2234 [ [ [ ] COCl] Cl][CO] Cl = + K K K 3 4 5 2 (2) \u2234 For steady state of Cl, d dt [ ] Cl = 0 or 2 K 1 [Cl 2 ] \u2013 2 K 2 [Cl] 2 \u2013 K 3 [Cl][CO] + K 4 [COCl] + K 5 [COCl][Cl 2 ] = 0 \u2234 [ [ ] / Cl]= Cl K K 1 2 2 1 2 \u239b \u239d \u239c \u239e \u23a0 \u239f (3) From (1), (2), (3), r K K K K K K COCl CO Cl 2 1 1 2 5 3 2 3 2 2 1 2 4 5 2 = + / / / [ ][Cl ] ( [ ])
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: r 4 >> r 5 or K 4 [COCl] >> K 5 [COCl][Cl 2 ] or K 4 >> K 5 [Cl 2 ] \u2234 r K K K K K K K K K COCl CO Cl CO 2 1 1 2 3 5 2 3 2 2 1 2 4 5 2 1 1 2 3 5 = + \u2248 / / / / [ ][Cl ] ( [ ]) [ ] ][Cl ] / / 2 3 2 2 1 2 4 K K
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: A overall = A A A A A 1 1 2 3 5 2 1 2 4 / / \u22c5 \u22c5 \u22c5
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: E E E E E E a a a a a a overall = + + \u2212 \u2212 1 2 1 2 1 3 5 2 4 Comprehension III For steady state of Br, + = d dt [Br] 0 or, 2 K 1 [Br 2 ] \u2013 K 2 [Br][H 2 ] + K 3 [H][Br 2 ] + K 4 [H][HBr] \u2013 2 K 5 [Br] 2 = 0 (1) For steady state of H, + = d dt [H] 0 or, K 2 [Br][H 2 ] \u2212 K 3 [H][Br 2 ] \u2013 K 4 [H][HBr] = 0 (2)
Comprehension 2 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: From (1) and (2), [ [ ] / Br] Br = \u239b \u239d \u239c \u239e \u23a0 \u239f K K 1 2 5 1 2
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: [ [Br][H ] [Br ] [HBr]) [Br ] [H ] / / / H] = + = \u22c5 \u22c5 \u22c5 K K K K K K 2 2 3 2 4 2 1 1 2 2 1 2 2 5 1 2 ( ( [ ] [ K K 3 2 4 Br HBr]) +
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: + d dt [HBr] = K 2 [Br][H 2 ] + K 3 [H][Br 2 ] \u2013 K 4 [H][HBr] = = + 2 2 3 2 3 2 1 1 2 2 3 2 2 5 1 2 3 2 4 K K K K K K K [ ] [Br ] [H ] ( [ ] [ / / / H][Br Br HBr])
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 580 in PDFExtracted text
Solution: At t = 0, [HBr] = 0 and hence, initial rate is given by, r K K K 0 2 1 1 2 2 1 2 2 5 1 2 2 = / / / [Br ] [H ]\n11.54 Chapter 11 HINTS AND EXPLANATIONS Comprehension IV
Comprehension 3 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: K t t V V t t = \u22c5 = \u22c5 1 1 0 0 ln ln [H O ] [H O ] 2 2 2 2 For t = 10 min, K 1 1 1 10 25 6 16 1 6 10 = \u22c5 = \u2212 ln . ln . min For t = 20 min, K 2 1 1 20 25 6 10 1 6 10 = \u22c5 = \u2212 ln . ln . min As K 1 = K 2 , order of reaction = 1
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: t K 1 2 2 2 1 6 10 15 / ln log log . min = = \u239b \u239d \u239c \u239e \u23a0 \u239f =
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: Kt a a x x a t = = \u2212 = \u2212 ln ln ln [H O ] [H O ] 2 2 2 2 0 1 1 or, ln . ln 1 6 10 25 1 1 \u00d7 = \u2212 x a \u21d2 x a = 11 16
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: Order = 1, but molecularity = 2 (as per reaction). Comprehension V C 8 H 18 O 2 ( g ) \u2192 2CH 3 COCH 3 ( g ) + C 2 H 6 ( g ) t = 0 800 torr 0 0 t = t (800 \u2013 x ) torr 2 x torr x torr
Comprehension 4 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: P t t C H O 8 18 2 torr torr = \u23af \u2192 \u23af\u23af\u23af = \u00d7 800 100 3 1 2 / \u2234 t = 3 \u00d7 80 = 240 min
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: P acetone = 2 x = 1200 \u21d2 x = 600 \u2234 P t t C H O 8 18 2 torr torr = \u23af \u2192 \u23af\u23af\u23af = \u00d7 800 200 2 1 2 / \u2234 t = 2 \u00d7 80 = 160 min
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: 800 \u2013 x = 700 \u21d2 x = 100 \u2234 P total = (800 \u2013 x ) + 2 x + x = 1000 torr Comprehension VI
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: From (1) and (2) data : order w.r.t OH = 1 From (2) and (3) data : order w.r.t H 2 S = 1 \u2234 r = K [H 2 S][OH]
Comprehension 5 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: K r = = \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 [ . ( . ) ( . ) H S][OH] M s M M 2 1 4 10 2 1 10 1 3 10 6 1 8 8 = 5.1 \u00d7 10 9 M \u20131 s \u20131
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: r = K [H 2 S][OH] = 5.1 \u00d7 10 9 \u00d7 (1.0 \u00d7 10 \u20138 ) \u00d7 (1.7 \u00d7 10 \u20138 ) = 8.67 \u00d7 10 \u20137 M s \u20131
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: r = 8.67 \u00d7 10 \u20137 \u00d7 0.1 = 8.67 \u00d7 10 \u20138 mol s \u20131 Comprehension VII
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: K t P P x x = \u22c5 \u00b0 1 ln For t = 100 min, K 1 1 1 100 800 400 2 100 = \u22c5 = \u2212 ln ln min For t = 200 min, K 2 1 1 200 800 200 2 100 = \u22c5 = \u2212 ln ln min As K 1 = K 2 , order of reaction = 1
Comprehension 6 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: K = = \u00d7 \u2212 \u2212 ln . min 2 100 6 93 10 3 1 \u2234 K K rxn = = \u00d7 \u2212 \u2212 2 3 465 10 3 1 . min
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: Time for 87.5 % reaction = 3 3 2 6 93 10 1 2 3 \u00d7 = \u00d7 \u00d7 \u2212 t / ln . = 300 min
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 581 in PDFExtracted text
Solution: 2X( g ) 3Y( g ) + 2Z( g ) t = 0 800 0 0 t = t 800 \u2013 x 3 2 x x = 700 \u2234 P total = 800 + 3 2 x = 950 torr\n11.55 Chemical Kinetics HINTS AND EXPLANATIONS Comprehension VIII
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: A + 2B C + D t = 0 a M b M 0 0 t = t ( a \u2013 x ) M ( b \u2013 2 x ) M Now, r = K \u22c5 C B \u21d2 \u2212 = \u2212 d dt K b x [A] ( ) 2 \u21d2 dx dt K b x = \u2212 ( ) 2 or, dx b x K dt x t \u2212 = \u22c5 \u222b \u222b 2 0 0 \u21d2 x b e Kt = \u2212 \u2212 2 1 2 ( ) \u2234 C A = a \u2013 x = a b e Kt \u2212 \u2212 \u2212 2 1 2 ( )
Comprehension 7 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: For C a a a b e A Kt = = \u2212 \u2212 \u2212 2 2 2 1 2 , ( ) \u2234 ( ) ln / t K b b a A 1 2 1 2 = \u22c5 \u2212
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: For ( ) ( ) , [ ] [ ] / / t t A B a b A B 1 2 1 2 1 2 = = = Comprehension IX
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: r n dn dt K n rxn A A = \u2212 \u22c5 = \u22c5 1 1 \u21d2 n A = n A \u00b0 \u22c5 e \u2013 n , kt
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: n 1 A n 2 A t = 0 a mole 0 t = t ( a \u2013 x )mole n n x 2 1 \u22c5 mole = a \u22c5 e \u2013 n , kt \u2234 x = a (1 \u2013 e \u2013 n , kt ) Now, V V n n a x n n x a 2 1 2 1 = = \u2212 + \u22c5 final initial ( ) or, V V a x n n a n n e n kt 2 0 2 1 2 1 1 1 1 1 = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u2212 \u2212 ( ) , \u2234 V V n n n n e n kt 2 0 2 1 2 1 1 = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 ,
Comprehension 8 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: If n 1 = 1, n 2 = 2, then V 2 = V 0 (2 \u2013 e \u2013 kt ) Now, [ ] ( ) [ ] A n V n e V e A e e A A kt kt kt kt = = \u00b0 \u22c5 \u22c5 \u2212 = \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u2212 \u2212 2 0 0 2 2 Comprehension X
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: df f K dt f t 1 0 0 \u2212 = \u22c5 \u222b \u222b \u21d2 t f K = \u2212 \u2212 ln( ) 1 Now, K = \u2212 \u2212 = \u2212 ( ) 3 200 3 200 1 hr \u2234 t K 1 2 2 0 693 3 200 46 2 / ln . . = = \u239b \u239d \u239c \u239e \u23a0 \u239f = hr
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: t f K = \u2212 \u2212 ln( ) 1 \u21d2 f = 1 \u2013 e \u2013 Kt = 1 \u2013 e \u20133 t /200 Comprehension XI
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: Unit of K = s \u20131 \u21d2 order = 1
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 582 in PDFExtracted text
Solution: K K B = \u00d7 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 10 100 10 100 1 5 10 1 5 10 4 5 1 . . s\n11.56 Chapter 11 HINTS AND EXPLANATIONS Comprehension XII
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: [ ] [ ] ( . ) . M . A A e e K t = \u22c5 = \u00d7 = \u2212 \u2212 \u00d7 0 0 04 25 1 1 0 0 368 M
Comprehension 10 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: [ ] [ ] ( ) B K A K K e e K t K t = \u2212 \u2212 \u2212 \u2212 1 0 2 1 1 2 = \u00d7 \u2212 \u2212 \u2212 \u00d7 \u2212 \u00d7 0 04 1 0 0 06 0 04 0 04 25 0 06 25 . ( . . . ( ) . . M) e e = 0.29 M
Comprehension 11 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: [ C ] = [ A 0 ] \u2013 [ A ] \u2013 [ B ] = 1.0 \u2013 0.368 \u2013 0.29 = 0.342 M
Comprehension 11 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: t K K K K max ln ln . . min = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = 2 1 2 1 3 2 0 06 0 04 20
Comprehension 11 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: [ ] [ ] ( . ) max . . . B A K K K K K = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u2212 \u2212 0 2 1 0 06 0 06 0 0 2 2 1 1 0 3 2 M 4 4 = 0.3 M
Comprehension 12 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: [ ] [ ] [ ] [ ] A B C A = = = 0 3 Now, t K A A K = \u22c5 = = = 1 3 1 1 0 04 27 5 1 0 1 ln [ ] [ ] ln . . . min
Comprehension 12 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: \u2212 = + d A dt d C dt [ ] [ ] \u21d2 K 1 [ A ] = K 2 [ B ] \u21d2 t = 20 min \u2234 [ ] [ ] ( . M) e . . A A e K t = \u22c5 = \u22c5 = \u2212 \u2212 \u00d7 0 0 04 20 1 1 0 0 45M
Comprehension 12 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: ( r C ) max = K 2 [ B ] max = 0.06 \u00d7 0.3 = 1.8 \u00d7 10 \u20132 M/ min Comprehension XIII A B t = 0 0.15 M 0 t = 10 (0.15 \u2013 x ) M x M = 0.125 M = 0.025 M t = t eq (0.15 \u2013 x eq ) M x eq M = 0.10 M = 0.05 M Now, K K K eq f b = = = 0 05 0 10 1 2 . . (1) and t K K f b 1 2 2 / ln = + \u21d2 10 0 693 min . = + K K f b (2)
Comprehension 13 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: From (1) and (2), K f = 2.31 \u00d7 10 \u20132 min \u20131
Comprehension 13 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: From (1), K b = 4.62 \u00d7 10 \u20132 min \u20131
Comprehension 13 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Comprehension 14 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: t 1/2 = 10 min Comprehension XIV
Comprehension 14 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 583 in PDFExtracted text
Solution: E K E K E K K a a a ( ) overall = \u22c5 + \u22c5 + 1 2 1 2 1 1 or, 10 5 12 9 1 2 1 2 . = \u00d7 + \u00d7 + K K K K \u21d2 K 1 = K 2 or, A e A e E RT E RT a a 1 2 1 2 \u22c5 = \u22c5 \u2212 \u2212 / / or, E E RT A A a a 1 2 1 2 \u2212 = ln \u21d2 ( ) ln 12 9 10 2 2 10 2 10 3 14 14 2 \u2212 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 T e \u2234 T = 750 K
Comprehension 14 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Above 750 K, Y will be the major product and below 750 K, Z will be the major product as E E a a 1 2 > .
Comprehension 15 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Reactions with higher E a are more sensitive towards temperature change.\n11.57 Chemical Kinetics HINTS AND EXPLANATIONS Comprehension XV Energy (kcal/mol) 27.5 19.9 30.1 16.9 67.9 10.2 2.4 4.3 13.2 7.6 42.8 A B C D Reaction coordinates
Comprehension 15 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: C \u2192 D [ : ] . . . E a A B C D 27 5 30 1 4 3 \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af
Comprehension 15 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: C \u2192 B [ : D A] . . . E a 67 9 16 9 19 9 \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af C B
Comprehension 16 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: C \u2192 D [Lowest E a ]
Comprehension 16 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: B \u2192 C [Highest E a ]
Comprehension 16 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: D \u2192 C [Highest E a ]
Answer: C
\nOriginal PDF solution page
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Solution: Molecularity can never be fractional.
Answer: C
\nOriginal PDF solution page
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Solution: t A K n n 100 0 1 1 % [ ] ( ) = \u2212 \u2212 when n < 1 = Infi nite when n \u2265 1
Answer: D
\nOriginal PDF solution page
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Solution: For a particular step, rates always increase with increase in temperature.
Answer: A
\nOriginal PDF solution page
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Solution: Relative increase in rate constant with increase in temperature is higher for the reaction with higher activation energy.
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 H = E E a a f b \u2212
Answer: A
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: For zero order reaction : t A K t A K 1 2 0 100 0 2 / % [ ] , [ ] = =
Answer: D
\nOriginal PDF solution page
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Solution: Order is in dependent from stoichiometry of reaction.
Answer: A
\nOriginal PDF solution page
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Solution: t A K 1 2 0 2 / [ ] =
Answer: A
\nOriginal PDF solution page
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Answer: A \u2192 P, Q; B \u2192 R, S; C \u2192 P, Q; D \u2192 R, S
\nOriginal PDF solution page
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Answer: A \u2192 Q; B \u2192 P; C \u2192 P; D \u2192 R, S
\nOriginal PDF solution page
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Answer: A \u2192 Q; B \u2192 P, R; C \u2192 P; D \u2192 R, S
\nOriginal PDF solution page
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Answer: A \u2192 R; B \u2192 P; C \u2192 Q, S; D \u2192 Q, S
\nOriginal PDF solution page
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Answer: A \u2192 P; B \u2192 S; C \u2192 R; D \u2192 Q
\nOriginal PDF solution page
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Answer: A \u2192 R, U; B \u2192 P, X; C \u2192 S, V; D \u2192 Q, W
\nOriginal PDF solution page
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Answer: A \u2192 Q; B \u2192 P, R; C \u2192 S; D \u2192 T
\nOriginal PDF solution page
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Answer: A \u2192 P; B \u2192 Q, S; C \u2192 R, T; D \u2192 U
\nOriginal PDF solution page
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Answer: A \u2192 S; B \u2192 P; C \u2192 R, T; D \u2192 Q
\nOriginal PDF solution page
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Solution: (P) 2 a a t 1/3 t 19/27 = 54 sec 3 = 18 sec t 1/3 = 18 sec t 1/3 = 18 sec 4 a 9 8 a 27 (Q) 3 a a t 1/4 t 7/16 = 32 sec 4 = 16 sec t 1/4 = 16 sec 9 a 16 (R) K a a a x a = \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u2212 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 1 4 1 2 3 1 1 56 1 1 / \u21d2 x a = 7 8\n11.58 Chapter 11 HINTS AND EXPLANATIONS (S) K a a x = \u2212 = 2 3 18 30 \u21d2 x a = 5 9 (T) K a a x = \u2212 = 2 16 28 \u21d2 x a = 7 8
Answer: A \u2192 P, Q; B \u2192 P, R; C \u2192 S
\nOriginal PDF solution page
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Solution: (A) d C dt K B [ ] [ ] = 2 For d C dt [ ] max \u239b \u239d \u239c \u239e \u23a0 \u239f , [ B ] should be maximum and hence t K K K K K = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = ln ln 2 1 2 1 1 2 (when K 2 = 2 K 1 ) Now, ( ) ln / t K A 1 2 1 2 = (B) Rate of formation of B is maximum at t = 0, at which [ B ] = [ C ] = 0 Now, [ B ] = [ C ] K A K K e e A K e K e K K K t K t K t K t 1 0 2 1 0 2 1 2 1 1 2 1 2 1 [ ] ( ) [ ] \u2212 \u2212 = \u2212 \u22c5 \u2212 \u22c5 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u2212 \u2212 \u2212 \u2212 \u23a5 \u23a5 or, e e K e K e K K t K t K t K t \u2212 \u2212 \u2212 \u2212 \u2212 = \u2212 \u22c5 \u2212 \u22c5 1 1 1 1 2 1 1 2 1 1 2 (when K 2 = 2 K 1 ) or, K e K e K K e K e K t K t K t K t 1 1 2 1 1 1 2 1 1 1 1 2 \u22c5 \u2212 \u22c5 = \u2212 \u22c5 \u2212 \u22c5 \u2212 \u2212 \u2212 \u2212 \u2234 t K = ln 2 1 (C) [A] = [B] [ ] [ ] ( ) A e K A K K e e K t K t K t 0 1 0 2 1 1 1 2 \u22c5 = \u2212 \u2212 \u2212 \u2212 \u2212 K K K e K K t 2 1 1 1 1 2 \u2212 = \u2212 \u2212 ( ) \u2234 t K K K K K = \u2212 \u22c5 \u2212 1 2 1 2 1 2 1 ln
Answer: 3
\nOriginal PDF solution page
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Answer: 2
\nOriginal PDF solution page
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Answer: 6
\nOriginal PDF solution page
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Solution: K K K BrO BrO Br \u2212 \u2212 \u2212 = = 3 1 2 3 \u2234 K a BrO M s 3 0 06 3 0 02 1 1 \u2212 = = = \u2212 \u2212 . . and K b Br M s \u2212 = = \u00d7 = \u2212 \u2212 2 3 0 06 0 04 1 1 . .
Answer: 5
\nOriginal PDF solution page
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Solution: 7 2 10 3600 2 10 15 1 8 2 . ( \u00d7 = \u00d7 \u00d7 \u2212 \u2212 \u2212 M s M) K K = \u2212 \u2212 1 200 1 1 M s = 5 ml mol \u20131 s \u20131
Answer: 3
\nOriginal PDF solution page
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Solution: \u2212 = \u22c5 \u22c5 dP dt K P P a b NO H 2 1 5 0 25 372 152 . . = \u239b \u239d \u239c \u239e \u23a0 \u239f a \u21d2 a = 2 and 1 60 0 79 289 144 . . = \u239b \u239d \u239c \u239e \u23a0 \u239f b \u21d2 b = 1
Answer: 1
\nOriginal PDF solution page
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Solution: 0 1 0 4 0 20 . . . = x \u21d2 x = 0.8 0 1 0 8 0 2 0 05 0 4 . . . . . = \u00d7 \u00d7 y \u21d2 y = 0.2
Answer: 5
\nOriginal PDF solution page
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Solution: t K = \u22c5 + + 1 3 0 3 ln [ ] [ ] Cr Cr = \u00d7 \u22c5 \u2212 \u2212 \u2212 1 9 10 100 100 80 5 1 s ln = 1.8 \u00d7 10 4 sec = 5 hrs
Answer: 3
\nOriginal PDF solution page
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Solution: (i) Addition of NaOH will decrease [H 3 O + ]. (ii) Addition of water will decrease the concentration of both. (iii) Acetic acid is a weak acid and hence, [H 3 O + ] will decrease. (iv) Increase in temperature increases the reaction rate.\n11.59 Chemical Kinetics HINTS AND EXPLANATIONS
Answer: 3
\nOriginal PDF solution page
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Solution: Time for certain progress of reaction, t a [ A 0 ] 1 \u2013 n 1 10 0 25 10 0 02 0 04 3 3 1 \u00d7 \u00d7 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u2212 . . . n \u21d2 n = 3
Answer: 2
\nOriginal PDF solution page
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Solution: C 4 H 8 2C 2 H 4 t = 0 a mole 0 t = t ( a \u2013 x ) mole 2 x mole As, a \u2013 x = 2 x \u21d2 x a = 3 Now, t t K a a x a a a = \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = \u2212 \u2212 ln ln 1 25 18 10 3 2 5 1 s hrs
Answer: 8
\nOriginal PDF solution page
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Solution: For 2 % reaction, we may assume that rate is almost constant. r = K [ A ] \u21d2 2 100 1 \u00d7 \u2212 [ ] min A = K [ A ] \u21d2 K = 0.02 min \u20131
Answer: 3
\nOriginal PDF solution page
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Solution: H 2 O 2 ( aq ) H 2 O( l ) + 1 2 O 2 ( g ) \u0394 H = (\u2013287) \u2013 (\u2013 187) = \u2013100 KJ/mol Moles of H 2 O 2 reacted per sec = 7.5 \u00d7 10 \u20134 \u00d7 0.02 \u00d7 2 = 3 \u00d7 10 \u20135 \u2234 Heat produced per sec = 3 \u00d7 10 \u20135 \u00d7 (100 \u00d7 10 3 ) = 3 J
Answer: 2
\nOriginal PDF solution page
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Solution: t K a a x = \u22c5 \u2212 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u2212 \u2212 1 1 4 5 3 1536 10 100 40 8 1 ln . . ln s = \u00d7 \u00d7 \u00d7 \u00d7 = 0 9 3 1536 10 4 5 3 1536 10 2 8 7 . . . . Year Years
Answer: 5
\nOriginal PDF solution page
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Solution: t 1 2 4 0 693 6 93 10 1000 / . . sec = \u00d7 = \u2212 A n B t = 0 a mole 0 t = 1000 sec a 2 mole n a \u22c5 2 mole Now, a n a a 2 2 3 + \u22c5 = \u21d2 n = 5
Answer: 2
\nOriginal PDF solution page
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Solution: r = K [ester][H + ] x = k 1 [ester] K 1 = K \u22c5 [H + ] x 1 0 10 10 10 3 2 . = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 x \u21d2 x = 1 and K K y 1 1 3 3 1 0 10 10 1 = = \u00d7 = = + \u2212 \u2212 [ ] . H
Answer: 2
\nOriginal PDF solution page
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Solution: t 3/4 = 2 \u00d7 t 1/2 and hence, a =
Answer: 1
\nOriginal PDF solution page
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Solution: Now, t K K b 1 2 2 2 / ln ln [ ] = = + H 1 0 0 5 0 02 0 01 . . . . = \u239b \u239d \u239c \u239e \u23a0 \u239f b \u21d2 b = 1
Answer: 4
\nOriginal PDF solution page
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Solution: Concentrations are in G.P. and hence, order =
Answer: 8
\nOriginal PDF solution page
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Solution: 18. A 2 B 3 ( aq ) 2A 3+ ( aq ) + 3 B 2\u2013 ( aq ) t = 0 a mole 0 0 t = 10 min a \u2013 x 2 x 3 x Now, p = CRT = r gh \u21d2 total mole a h \u2234 a a x + = 4 2 6 \u21d2 t = 10 min = t 1/2 Now, at t = t 3/4 = 2 \u00d7 t 1/2 = 20 min, x a = 3 4 \u2234 a a a h + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 4 3 4 2 \u21d2 h = 8 mm p = x = r gh = 1 0 1000 0 8 3 2 . ( . gm cm cm s cm) \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = = 800 80 2 dyne cm pascal Now, x y = = 80 20 4
Answer: 0
\nOriginal PDF solution page
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Solution: [ ] A t 4 1 1 = + \u21d2 4 1 1 3 2 8 [ ] [ ] ( ) [ ] A d A dt t A \u22c5 = \u2212 + = \u2212 \u2234 \u2212 = = = \u00d7 \u2212 \u2212 d A dt A [ ] [ ] ( . ) 5 5 5 1 4 0 2 4 8 10 M s
Answer: 8
\nOriginal PDF solution page
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Solution: A 2B t = 0 a mole 0 t = t ( a \u2013 x ) mole 2 x mole From mass conservation, a \u00d7 M 0 = ( a + x ) \u00d7 M t \u2234 x a M M M t t = \u2212 ( ) 0 If the reaction is zero order, then K a a x t x t a M M t M t t = \u2212 \u2212 = = \u2212 \u22c5 ( ) ( ) 0\n11.60 Chapter 11 HINTS AND EXPLANATIONS For t = 10 min, K a a = \u2212 \u00d7 = ( ) 42 35 10 35 50 For t = 20 min, K a a = \u2212 \u00d7 = ( ) 42 30 20 30 50 As K values are same, the reaction is of zero-order.
Answer: 2
\nOriginal PDF solution page
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Solution: r = K [ A ] n and 2 r = K (4[ A ]) n \u21d2 n = 1 2 \u2234 t 1/2 a [ A 0 ] 1 \u2013 n = [ A 0 ] 1/2 Next t 1/2 will be 1 2 times of previous one and hence, t = = 8 2 2 8 hr.
Answer: 1
\nOriginal PDF solution page
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Solution: P K P K K K e B B A A B C K K K t A B C = \u22c5 \u00b0 + + \u22c5 \u2212 \u2212 + + \u22c5 [ ] ( ) 1 = \u00d7 \u00d7 \u00d7 \u2212 = \u2212 \u2212 \u2212 \u00d7 \u00d7 \u2212 2 10 13 86 6 93 10 1 2 3 3 6 93 10 100 3 . . [ ] . e atm
Answer: 7
\nOriginal PDF solution page
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Solution: K K A e A e A A e E RT E RT E E RT a a a a I II I II I II I II I II = \u22c5 \u22c5 = \u22c5 \u2212 \u2212 \u2212 \u2212 / / ( )/ = = \u00d7 = \u2212 \u00d7 \u00d7 100 1 4 606 10 2 500 3 e . /
Answer: 4
\nOriginal PDF solution page
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Solution: Fraction of molecules having sufficient energy = = = \u00d7 \u2212 \u2212 \u00d7 \u00d7 \u2212 e e E RT a / . / . 83 14 10 8 314 500 9 3 2 10
Answer: 7200
\nOriginal PDF solution page
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Solution: ln ln K K t t E R T T a 2 1 1 2 1 2 1 1 = = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, ln 1 3 1 300 1 280 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E R a (1) and ln 16 1 300 1 330 t E R a = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f (2) From (1) and (2), t = 4 hrs Four Digit Integer Type
Answer: 0025
\nOriginal PDF solution page
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Solution: r = K [O 3 ] 2 = 5 \u00d7 10 \u20134 \u00d7 (2 \u00d7 10 \u20138 ) 2 = 2 \u00d7 10 \u201319 mol l \u20131 s \u20131 = 2 \u00d7 10 \u201319 \u00d7 6 \u00d7 10 23 \u00d7 10 \u20133 \u00d7 60 = 7200 molecules ml \u20131 min \u20131
Answer: 0060
\nOriginal PDF solution page
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Solution: At t = \u221e , P total should be 400 mm, but as it is only 390 mm, some unreactive gas should also be present in the vessel. Let P P \u00b0 = C H Br 2 5 0 mm then P unreactive gas = (200 \u2013 P 0 ) mm. C 2 H 5 Br(g) C 2 H 4 (g) + HBr(g) t = 0 P 0 0 0 t = t P 0 \u2013 x x x t = \u221e 0 P 0 P 0 From question, P 0 + P 0 + (200 \u2013 P 0 ) = 390 \u21d2 P 0 = 190 and ( P 0 \u2013 x ) + x + x + (200 \u2013 P 0 ) = 342.5 \u21d2 x = 142.5 \u2234 Percentage C 2 H 5 Br undecomposed = P x P 0 0 25 \u2212 = %
Answer: 1250
\nOriginal PDF solution page
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Solution: t t A A = \u22c5 gen log log [ ] [ ] 2 0 \u21d2 96 0 30 3 = \u22c5 t gen . log \u21d2 t gen = 60 hrs
Answer: 1000
\nOriginal PDF solution page
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Solution: r dP dt K P P = \u2212 = \u2032 \u22c5 \u22c5 NO NO O 2 2 and \u2032 = \u00d7 \u00d7 \u2212 \u2212 K 1 6 10 0 08 600 5 2 2 1 . ( . ) atm s \u2234 r = \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 6 10 48 48 190 760 288 760 5 2 . = = \u2212 \u2212 1250 760 1250 1 1 atm s mm s
Answer: 0200
\nOriginal PDF solution page
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Solution: From the unit of rate constant, the process is zero order. \u2234 t K 100 0 % [ ] = + H = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = \u00d7 = \u2212 \u2212 3 10 0 05 1000 1 0 10 6 10 1000 7 7 4 . . sec min
Answer: 0040
\nOriginal PDF solution page
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Solution: K t A A = = 1 0 ln [ ] [ ] Constant \u2234 1 100 3 1 9 0 0 0 0 \u22c5 = \u22c5 ln [ ] [ ] / ln [ ] [ ] / A A t A A \u21d2 t = 200 min
Answer: 0120
\nOriginal PDF solution page
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Solution: K t A A = \u22c5 = 1 0 ln [ ] [ ] Constant\n11.61 Chemical Kinetics HINTS AND EXPLANATIONS \u2234 1 20 500 420 1 100 70 \u22c5 = \u22c5 ln ln t \u21d2 t = 40 min
Answer: 0535
\nOriginal PDF solution page
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Solution: K t V V V t = \u22c5 \u2212 = \u221e \u221e 1 ln Constant \u2234 1 40 80 80 40 1 80 80 70 \u22c5 \u2212 = \u22c5 \u2212 ln ln t \u21d2 t = 120 min
Answer: 0123
\nOriginal PDF solution page
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Solution: 2P 4Q + R + S(l) t = 0 P 0 0 0 t = 30 min P 0 \u2013 x 2 x x 2 V.P. = 25 t = 60 min P 0 \u2013 y 2 y y 2 V.P. = 25 t = \u221e 0 2 P 0 P 0 2 V.P. = 25 From question, 2 2 25 625 0 0 P P + + = \u21d2 P 0 = 240 and ( ) P x x x 0 2 2 25 445 \u2212 + + + = \u21d2 x = 120 Now, 1 30 1 60 0 0 0 0 \u22c5 \u2212 = \u22c5 \u2212 ln ln P P x P P y \u21d2 y = 180 \u2234 P P y y y 60 0 2 2 25 535 = \u2212 + + + = ( ) mm
Answer: 0375
\nOriginal PDF solution page
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Solution: Initial moles of NH 4 NO 2 = 200 0 02 1000 0 004 \u00d7 = . . and moles of N 2 O formed = ( ) . . . 785 25 760 49 26 1000 0 0821 300 0 002 \u2212 \u00d7 \u00d7 = \u2234 t req = t 1/2 = 123 min
Answer: 0011
\nOriginal PDF solution page
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Solution: For set 1 and 2, r = K \u2032 [ B ] as [ A 0 ] >> [ B 0 ] and t K K A 1 2 0 2 2 2 / ln ln [ ] = \u2032 = \u21d2 x = 62.5 For set 3 and 4, r = K \u2033 [ A ] 2 as [ B 0 ] >> [ A 0 ] and t K A K B A 1 2 0 0 0 1 1 / [ ] [ ][ ] = \u2032\u2032 = \u21d2 y = = 625 2 312 5 . \u2234 x + y = 62.5 + 312.5 = 375
Answer: 1680
\nOriginal PDF solution page
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Solution: t = 43.5 min = 3 t 1/2 Hence, P ether atm = = 4 2 0 5 3 . \u21d2 \u0394 P ether = 3.5 atm \u2234 P fi nal = 0.5 + 3.5 \u00d7 3 = 11 atm
Answer: 4003
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 t t r r 1 2 1 2 2 / ln ln \u21d2 12 2 0 04 0 03 1 2 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f t / ln ln . . \u2234 t 1/2 = 28 min = 1680 sec
Answer: 0025
\nOriginal PDF solution page
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Solution: t t V V V V t = \u22c5 \u2212 \u2212 \u221e \u221e 1 2 0 2 / ln ln \u21d2 120 2 60 20 60 55 1 2 = \u22c5 \u2212 \u2212 t / ln ln \u2234 t 1/2 = 40 min ab = 40 Now, [ [ester] HCl] = \u2212 \u221e V V V 0 0 \u21d2 [ . HCl] 6 0 20 60 20 = \u2212 \u21d2 [HCl] = 3.0 M \u2234 cd = 03
Answer: 0500
\nOriginal PDF solution page
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Solution: t 1 2 3 0 693 1 386 10 500 / . . sec = \u00d7 = \u2212 Let the initial moles of A = x , then after 500 sec, A 2B + C x x \u2212 2 2 2 \u00d7 x x 2 = x 2 = x = x 2 Total moles becomes x x x x 2 2 2 + + \u239b \u239d \u239c \u239e \u23a0 \u239f = . As moles becomes double, volume becomes double and hence, [ ] . . A req M = \u00d7 = 0 1 2 2 0 025 = 25 millimole per litre
Answer: 0260
\nOriginal PDF solution page
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Solution: A 2B + C t = 0 4 a 0 3 a t = t 4 a \u2013 x 2 x 3 a + x From question (4 a \u2013 x )(40\u00b0) + 2 x (10\u00b0) + (3 a + x ) (\u201330\u00b0) = 0\u00b0 \u2234 x a = 7 5 Now, t K a a a = \u22c5 \u2212 = \u22c5 = 1 4 4 7 5 1 0 001 20 13 500 ln . ln min\n11.62 Chapter 11 HINTS AND EXPLANATIONS
Answer: 0060
\nOriginal PDF solution page
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Solution: Exp (1): r = K \u2032 [ B ] y as [ A 0 ] >> [ B 0 ] \u2235 t t 7 8 1 2 3 / / = \u00d7 \u21d2 y = 1 Exp (2): r = K \u2033 [ A ] x as [ A 0 ] << [ B 0 ] \u2235 t t 7 8 1 2 7 / / = \u00d7 \u21d2 x = 2 Now, for exp (2) and (3), t K A K B A 1 2 0 0 0 1 1 / [ ] [ ][ ] = \u2032\u2032 = \u2234 a = \u00d7 = 10 2 2 2 5 . and b = 7 \u00d7 2.5 = 17.5 And for exp (1) and (4), t K K A 1 2 0 2 2 / ln ln [ ] = \u2032 = \u2234 c = 30 \u00d7 2 = 60 and d = 3 \u00d7 60 = 180
Answer: 1595
\nOriginal PDF solution page
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Solution: Percentage yield = K K K 2 1 2 100 4 8 3 2 4 8 100 60 + \u00d7 = + \u00d7 = . . . %
Answer: 1475
\nOriginal PDF solution page
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Solution: A + 2B + 3C D t = 0 1.0 M 1.0 M 1.0 M 0 t = t 1 \u2013 x 1 \u2013 2 x 1 \u2013 3 x x = 0.9 = 0.8 = 0.7 = 0.1 (given) \u2234 r = 2 \u00d7 10 \u20136 \u00d7 (0.9) 2 \u2013 1 4 10 0 1 0 8 0 7 1 595 10 6 2 6 . ( . ) . . . \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212
Answer: 0100
\nOriginal PDF solution page
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Solution: [ C ] = 0.875 + 0.6 = 1.475 M
Answer: 0055
\nOriginal PDF solution page
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Solution: K K K B A eq f b = = [ ] [ ] \u21d2 1 38 300 0 1 0 2 . / . . K b = \u21d2 K b = \u2212 1 38 150 1 . min Now, t K K x x x f b e B e B B = + \u22c5 \u2212 = + \u22c5 \u2212 \u00d7 1 1 1 38 300 2 76 300 0 1 0 1 0 3 25 100 ln . . ln . . . , , = \u00d7 \u22c5 = 300 6 2 4 100 ln ln min
Answer: 0727
\nOriginal PDF solution page
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Solution: For completion in 30 min, the rate should be increased by 4 60 30 8 \u00d7 = times. Assuming temperature coefficient constant, the approximate temperature is 25 10 8 2 55 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00b0 C .
Answer: 0500
\nOriginal PDF solution page
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Solution: K 1 = K 2 \u21d2 A e A e E RT E RT a a 1 2 1 2 \u22c5 = \u22c5 \u2212 \u2212 / / \u2234 ln A A E E RT a a 2 1 2 1 = \u2212 ln ( . . ) . 10 10 171 39 152 30 10 8 3 14 13 3 = \u2212 \u00d7 \u00d7 T or, T = 1000 K = 727\u00b0 C
Answer: 0100
\nOriginal PDF solution page
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Solution: t K A e E RT a 1 2 2 2 / / ln ln = = \u22c5 \u2212 or, 1 60 0 7 5 10 13 149 4 10 8 3 3 \u00d7 = \u00d7 \u00d7 \u2212 \u00d7 \u00d7 . . / . e T \u2234 T = 500 K
![\"Electrochemistry](\"./assets/chapter-packs/electrochemistry/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: (I) Cu cannot reduce Pb (II) Pb can reduce Ag (III) Ag cannot reduce Cu. Hence, reducing power: Pb > Cu > Ag
Answer: D
\nOriginal PDF solution page
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Solution: E E n = \u00b0 \u2212 \u22c5 0 06 . log [R] [O] \u21d2 0 24 0 36 0 06 1 . . . log [ [ = \u2212 R] O] \u2234 [ [ O] R] = 1 100
Answer: B
\nOriginal PDF solution page
Open page 433 in PDFExtracted text
Solution: For the complex ion to get oxidised, its reduction potential should be low.
Answer: B
\nOriginal PDF solution page
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Solution: Ag NH Ag(NH) E + + + \u00b0 = \u2212 = 2 0 79 0 37 0 42 3 3 \u001e \u21c0 \u001e \u21bd \u001e \u001e ; . . . V Now, E n K eq \u00b0 \u2212 \u22c5 0 06 . log \u21d2 0 42 0 06 1 . . log = \u22c5 K f \u21d2 K f = 10 7
Answer: A
\nOriginal PDF solution page
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Solution: Given: Co 3+ + e \u2013 Co 2+ ; E\u00b0 = 1.81 V; \u0394 \u00b0 = \u2212 \u00d7 \u00d7 G F 1.81 1 1 Co(CN) e Co(CN) 6 3 6 4 + \u2212 \u2212 + \u23af \u2192 \u23af E\u00b0 = \u20130.83 V; \u0394 \u00b0 = \u2212 \u00d7 \u00d7 \u2212 G F 2 1 0 83 ( . ) Co 6CN Co(CN) 2 6 4 + \u2212 \u2212 + \u23af \u2192 \u23af ; K f = 10 19 ; \u0394 \u00b0 = \u2212 \u00d7 G RT ln10 3 19 Required Co CN Co(CN) 3 6 3 6 + \u2212 \u2212 + \u23af \u2192 \u23af K f = ?; \u0394 \u00b0 = \u2212 \u00d7 G RT lnK f Now, \u0394 \u00b0 = \u0394 \u00b0 \u2212 \u0394 \u00b0 + \u0394 \u00b0 G G G G 1 2 3 or, \u2212 \u22c5 = \u2212 \u2212 + \u2212 RT lnK F) F RTln10 f ( . . ( ) 1 81 0 83 19 or, RT ln K F \u22c5 = \u2212 10 2 64 19 f . \u21d2 log . . . . 10 2 64 2 303 2 64 0 06 19 K F RT f = \u2212 = \u2212 \u2234 K f = 10 63
Answer: A
\nOriginal PDF solution page
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Solution: E E Cu E Cu Cu Cu Cu Cu Cu Cu 2 2 2 0 06 2 1 0 03 2 2 + + + = \u00b0 \u2212 \u22c5 = \u00b0 + + + | | | . log [ ] . log[ ] ] = 0.34 + 0.03 \u00d7 log(0.1) = 0.31 V \u2234 E Cu/Cu 2+ = \u2212 0.31 V
Answer: D
\nOriginal PDF solution page
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Solution: E E RT F Cd Ag cell cell = \u00b0 \u2212 \u22c5 + + 2 2 2 ln [ ] [ ] As CN\u2013 will form complex with Ag+ ion in the cathodic compartment, E cell will decrease.
Answer: D
\nOriginal PDF solution page
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Solution: \u2212\u0394 = \u22c5 = \u00b0 \u2212 \u22c5 = \u00b0 \u2212 + + G nF E nF E RT nF Zn Cu nFE cell cell Cell Cell [ ln [ ] [ ] 2 2 RT T C C \u22c5 ln 1 2
Answer: B
\nOriginal PDF solution page
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Solution: E E RT F K X Ag X|Ag Ag |Ag sp \u00b0 = \u00b0 \u2212 \u22c5 \u2212 + | ln 1
Answer: C
\nOriginal PDF solution page
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Solution: Cell reaction: H + (cathode, P H = 3) H + (anode, P H = ?) E H anode H P P cell anode H catho = \u2212 \u22c5 = \u2212 + + 0 0 059 1 0 059 . log [ ] [ ] cathode . d de H \u23a1 \u23a3 \u23a4 \u23a6 or, 0 272 0 059 3 . . [P ] = \u2212 anode H \u21d2 P anode H = 7.6
Answer: C
\nOriginal PDF solution page
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Solution: Cl 2 + H 2 O \u001f Cl \u2013 + ClO \u2013 + 2H + ; E V cell \u00b0 = \u2212 = \u2212 1 36 1 63 0 27 . . . Now, E E H cell cell = \u00b0 \u2212 + 0 06 1 2 . log[ ] or, 0 0 27 0 06 1 2 2 25 = \u2212 + \u00d7= . . . P H
Answer: A
\nOriginal PDF solution page
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Solution: E E E cell quinohydrone calomel = \u2212 0.210 = E quinohydrone \u2013 0.279 \u21d2 E quinohydrone = 0.489 V EXERCISE II (JEE ADVANCED)\n8.41 Electrochemistry HINTS AND EXPLANATIONS Quinohydrone electrode is + 2H + + 2e \u2013 (Quinone, Q) (Hydroquinone, H2Q) O O OH OH E E H E P H = \u00b0 \u2212 \u22c5 = \u00b0 \u2212 + 0 06 2 1 0 06 . log [ ] . or, 0.489 = 0.699 \u2013 0.06.P H \u21d2 P H = 3.5
Answer: A
\nOriginal PDF solution page
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Solution: Anode: Ag(s) Ag + (aq) + e \u2013 1 \u00d7 6 Cathode: Cr O aq H aq Cr aq H O(l) 2 7 2 3 2 14 6 2 7 \u2212 + \u2212 + + + \u23af \u2192 \u23af + ( ) ( ) e ( ) Net: 6 14 6 2 7 2 3 2 Ag(s) Cr O aq H aq Ag aq) 2Cr aq)+7H O(l) + + \u23af \u2192 \u23af + \u2212 + + + ( ) ( ) ( ( E E n Ag Cr Cr O H cell cell = \u00b0 \u2212 \u22c5 + + \u2212 + 0 06 6 3 2 2 7 2 14 . log [ ] [ ] [ ][ ] = (1.33 \u2013 0.80) \u2013 0 06 6 0 1 0 4 1 6 0 1 0 46 6 2 14 . log ( . ) ( . ) . ( . ) . \u22c5 \u00d7 \u00d7 = V
Answer: B
\nOriginal PDF solution page
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Solution: Net cell reaction: Zn \u2013 Hg(C 1 M) Zn \u2013 Hg(C 2 M) E C C V cell = \u2212 = \u2212 = 0 0 059 2 0 059 2 1 10 0 0295 2 1 . log . log .
Answer: B
\nOriginal PDF solution page
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Solution: E K cell eq \u00b0 = \u22c5 0 06 2 . log \u21d2 0 75 1 50 3 1 68 1 3 1 0 03 . . . . log \u2212 \u00d7 \u2212 \u00d7 \u2212 = K eq \u2234 K eq = 10 \u201322
Answer: B
\nOriginal PDF solution page
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Solution: [ ] . . H Ka C C M left 1 + \u2212 = \u22c5 = \u00d7 \u00d7 = 1 8 10 0 1 5 [ ] . . H K Kb C C M right w + \u2212 \u2212 = \u22c5 = \u00d7 \u00d7 = 10 1 8 10 0 01 14 5 2 Net cell reaction, assuming as concentration cell: H + (C 2 M) H + (C 1 M) E C C V cell = \u2212 \u22c5 = \u2212 0 0 06 1 0 465 1 2 . log .
Answer: A
\nOriginal PDF solution page
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Solution: E V V V \u00b0 = \u00d7 + \u00d7 \u2212 \u00d7 = + + 2 3 1 0 616 1 0 439 1 0 799 0 256 | . . . . \u2234 E V V V \u00b0 = \u2212 + + 3 2 0 256 | .
Answer: B
\nOriginal PDF solution page
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Solution: E E n H P cell cell H = \u00b0 \u2212 \u22c5 + 0 06 2 2 . log [ ] or, 0 70 0 28 0 0 06 2 1 2 . ( . ) . log [ ] = \u2212 \u2212 \u22c5 + H \u21d2 P H = 7.0
Answer: D
\nOriginal PDF solution page
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Solution: Net cell reaction: Ag C M) Ag C K M sp + + = \u23af \u2192 \u23af = \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f ( . / 1 2 1 3 0 1 2 4 Now, E C C cell = \u2212 \u22c5 0 0 06 1 2 1 . log \u21d2 0.162 = \u2212 \u22c5 0 06 2 0 1 1 3 . log ( ) . / K sp \u2234 K sp = 4 \u00d7 10 \u201312
Answer: A
\nOriginal PDF solution page
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Solution: E E V Tl Tl Pb Pb + + \u2212 = \u2212 | | . 2 0 444 or, E E Pb Tl Tl Tl Pb Pb + + \u2212 ( ) \u2212 \u22c5 = \u2212 + + | | . log [ ] [ ] . 2 0 06 2 0 444 2 2 or, [( . ) ( . )] . log . . . \u2212 \u2212 \u2212 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 0 336 0 126 0 03 0 1 0 1 0 444 2 K sp \u2234 K sp = 4 \u00d7 10 \u20136
Answer: B
\nOriginal PDF solution page
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Solution: E cell = E Cu \u2013 E Zn = (E Cu \u2013 E calomel ) \u2013 (E Zn \u2013 E calomel ) From question E calomel \u2013 E Zn = 1.083 V and E calomel \u2013 E Cu = \u2013 0.018 V
Answer: C
\nOriginal PDF solution page
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Solution: The potential of hydrogen electrode at H 2 (1 bar) may be expressed as E = \u2013 0.059 P H Now, E P Ka 1 0 059 = \u2212 + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 . log x y and E P Ka 2 0 059 = \u2212 + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 . log y x \u2234 P E E ) Ka 1 2 = \u2212 + ( . 0 118\n8.42 Chapter 8 HINTS AND EXPLANATIONS
Answer: A
\nOriginal PDF solution page
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Solution: For the given reaction, E E E cell given hydrogen \u00b0 = \u00b0 \u2212 \u00b0 \u2234 (\u20130.84) \u2013 0 = \u22c5 0 06 1 . log K eq \u21d2 K eq = 10 \u201314
Answer: A
\nOriginal PDF solution page
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Solution: Net cell reaction may be written as H 2 + Zn 2+ \u001f 2H + + Zn E E n H P cell cell H = \u00b0 \u2212 \u22c5 + + 0 06 2 2 2 . log [ ] [Zn ] or, ( . ) ( . ) . log [ ] . \u2212 = \u2212 \u2212 \u22c5 \u00d7 + 0 61 0 76 0 06 2 1 0 4 2 H \u2234 [H + ] = 2 \u00d7 10 \u20133 M Now, K H SO HSO a 2 3 2 3 3 2 2 10 6 4 10 0 4 = = \u00d7 \u00d7 \u00d7 + \u2212 \u2212 \u2212 \u2212 [ ][ ] [ ] ( ) ( . ) . = 3.2 \u00d7 10 \u20134
Answer: B
\nOriginal PDF solution page
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Solution: Cu NH Cu(NH K f 2 3 3 4 2 12 4 10 + + + = \u001e \u21c0 \u001e \u21bd \u001e \u001e ) , 1.0 M excess 100 % 0 1.0 M Equ. x 2.0 M 1.0 M 10 1 0 2 0 12 4 = \u00d7 . ( . ) x \u21d2 x = = \u00d7 \u2212 \u2212 10 16 6 25 10 12 14 . Now, cell reaction: Zn + Cu 2+ \u001f Zn 2+ + Cu E E Zn Cu cell cell = \u00b0 \u2212 \u22c5 + + 0 06 2 2 2 . log [ ] [ ] = \u2212 \u2212 \u22c5 \u00d7 = \u2212 [ . ( . )] . log . . . 0 34 0 76 0 06 2 1 0 6 25 10 0 704 14 V
Answer: C
\nOriginal PDF solution page
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Solution: Net cell reaction: Zn + 2H + \u001f Zn 2+ + H 2 E E Zn P H cell cell H = \u00b0 \u2212 \u22c5 + + 0 06 2 2 2 2 . log [ ] [ ] or, 0 70 0 0 76 0 06 2 0 01 1 2 . [ ( . )] . log . [ ] = \u2212 \u2212 \u2212 \u22c5 \u00d7 + H \u21d2 [H + ] = 0.01 M Moles of HCl in RHS = 500 0 01 1000 5 10 3 \u00d7 = \u00d7 \u2212 . \u2234 Mass of NaOH needed = 5 \u00d7 10 \u20133 \u00d7 40 = 0.2 gm
Answer: D
\nOriginal PDF solution page
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Solution: On assuming concentration cell, the net cell reaction is Ag + (C 1 M, Right) Ag + (C 2 M, left) Now, C M 1 0 1 40 100 0 04 = \u00d7 = . . and C K Cl K K M sp sp sp 2 0 1 50 100 0 05 = = \u00d7 = \u2212 [ ] . . Now, E C C cell = \u2212 0 0 06 1 2 1 . log or, 0 42 0 06 1 0 05 0 04 . . log / . . = \u2212 K sp \u21d2 K sp = 2 \u00d7 10 \u201310
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 G cell = \u2013 nFE cell \u21d2 \u2013 965 \u00d7 3 \u00d7 10 3 = \u201312 \u00d7 96500 \u00d7 E cell \u2234 E cell = 2.5 V
Answer: B
\nOriginal PDF solution page
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Solution: Theoretical efficiency = \u2212\u0394 \u00b0 \u2212\u0394 \u00b0 G H \u21d2 0 84 285 . = \u2212\u0394 \u00b0 G \u2234 \u0394 G\u00b0 = \u2013 0.84 \u00d7 285 KJ = \u2013nF \u22c5 E\u00b0 cell or, 0.84 \u00d7 285 \u00d7 10 3 = 2 \u00d7 96500 \u00d7 E\u00b0 cell \u21d2 E\u00b0 cell = 1.24 V
Answer: B
\nOriginal PDF solution page
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Solution: Net cell reaction: Ag(s) + H + + Cl \u2013 \u001f AgCl(s) + 1 2 H 2 (g) But for E cell calculation, reaction may be written as Ag(s) + H + \u001f Ag + + 1 2 H 2 Now, E E Ag P H cell cell H = \u00b0 \u2212 \u22c5 \u22c5 + + 0 06 1 2 1 2 . log [ ] [ ] / = \u2212 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = \u2212 \u2212 [ . ] . log . . . / 0 0 80 0 06 1 10 0 1 1 0 1 0 32 10 1 2 V It means that actual reaction is in reverse direction and E cell = 0.32 V
Answer: B
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: x y y x Ag NH Ag(NH + + + 3 3 \u001e \u21c0 \u001e \u21bd \u001e \u001e ) aM bM 0 b >> a Eqn. ? bM a x M K / Ag f x y a x b = \u22c5 + [ ] \u21d2 [Ag ] / + = \u22c5 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f a x b f y x K 1 Cell reaction: Ag + (Right) Ag + (Left)\n8.43 Electrochemistry HINTS AND EXPLANATIONS \u2234 E cell = 0 0 059 1 \u2212 \u22c5 + + . log [ ] [ ] Right Ag Left Ag Case-I: 0 118 0 059 4 10 4 10 4 2 1 . . log / = \u2212 \u22c5 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 x \u21d2 x = 1 Case-II: 0 118 0 059 0 1 1 . . log . y/ = \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f x \u21d2 y = 2
Answer: C
\nOriginal PDF solution page
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Solution: E K V cell eq \u00b0 = \u22c5 = \u00d7 = \u2212 0 06 1 0 06 1 1 667 10 0 3732 6 . log . log . . Now, E E V Cu Cu Cu Cu \u00b0 \u2212 \u00b0 = \u2212 + + + 2 0 3732 | | . (1) and E E E V Cu Cu Cu Cu Cu Cu \u00b0 = \u00d7 \u00b0 + \u00d7 \u00b0 + = + + + + 2 2 1 1 1 1 0 3376 | | | . or, E E Cu Cu Cu Cu \u00b0 + \u00b0 = + + + 2 0 6752 | | . V (2) From (1) and (2), E V Cu Cu \u00b0 = + | . 0 5242
Answer: A
\nOriginal PDF solution page
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Solution: Zn + Ni 2+ \u001f Zn 2+ + Ni E K cell eq \u00b0 = 0 06 2 . log \u21d2 (\u20130.24) \u2013 ( \u20130.75) = 0.03 log K eq \u2234 K eq = 10 17 \u21d2 It means that Ni 2+ will react almost completely and [Zn 2+ ] \u2248 1.0 M Now, 10 1 0 17 2 = + . [Ni ] \u21d2 [Ni 2+ ] = 10 \u201317 M
Answer: B
\nOriginal PDF solution page
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Solution: Assuming the cell as concentration cell, the cell reaction may be written as Ag C M Ag C M + \u2212 \u2212 + \u2212 \u2212 = \u00d7 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23af \u2192 \u23af = \u00d7 = 1 13 10 2 10 9 4 10 0 001 4 10 2 10 0 2 10 . . \u239b \u239b \u239d \u239c \u239e \u23a0 \u239f Now, E V cell = \u2212 \u22c5 \u00d7 = \u2212 \u2212 \u2212 0 0 06 1 10 4 10 0 024 9 10 . log .
Answer: A
\nOriginal PDF solution page
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Solution: 0 03 0 06 2 0 3 0 5 2 2 2 . . log [ ] [ ] . log . [ ] lower = \u22c5 = \u22c5 + + + Cu Cu Cu higher lower r \u2234 [Cu 2+ ] lower = 0.05 M
Answer: D
\nOriginal PDF solution page
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Solution: Cell reaction: H + (C 1 , HA 1 ) H + (C 2 , HA 2 ) E C C Ka Ka cell = \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 0 0 059 1 0 059 2 1 2 1 . log . log = \u2212 = 0 059 2 0 059 1 2 . ( ) . V P P K K a a
Answer: A
\nOriginal PDF solution page
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Solution: Au + + 2CN \u2013 \u001f Au ( ) CN 2 \u2212 , \u0394 G\u00b0 1 = \u2013RT \u22c5 ln x O 2 + 2H 2 \u039f + 4e \u2013 \u001f 40H \u2013 ; \u0394 G\u00b0 2 = \u20134 \u00d7 F \u00d7 0.41 Au 3+ + 3e \u2013 \u001f Au; \u0394 G\u00b0 3 = \u20133 \u00d7 F \u00d7 1.50 Au 3+ + 2e \u2013 \u001f Au + ; \u0394 G\u00b0 4 = \u20132 \u00d7 F \u00d7 1.40 From \u0394 \u00b0 + \u0394 \u00b0 \u2212 \u0394 \u00b0 + \u0394 \u00b0 \u0394 \u00b0 = \u2212 + G G G G G RT F required 1 2 3 4 1 4 1 29 , ln . x
Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: P H of right electrode will increase due to formation of OH \u2013 ion.
Answer: A
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: At low [Cl \u2013 ], O 2 becomes anode product
Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Solution: In the electrolysis of aq. KNO 3 , neither K + are NO 3 \u2212 participate in electrode reaction.
Answer: C
\nOriginal PDF solution page
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Solution: Na will react with water. S will not conduct electricity.
Answer: B
\nOriginal PDF solution page
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Solution: w E Q F = \u21d2 0 635 63 5 2 0 965 3600 96500 . . . \u00d7 = \u00d7 \u00d7 i \u21d2 i = 2 3 6 . A \u2234 % . . . % error = \u2212 = 2 3 6 0 5 2 3 6 10
Answer: B
\nOriginal PDF solution page
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Solution: Detonating gas is mixture of H 2 and O 2 n eq S n = n eq H 2 = n eq O 2 \u21d2 1 2 120 2 2 4 2 2 . \u00d7 = \u00d7 = \u00d7 n n H O \u2234 n and n H O 2 1 0 01 0 005 = = . .\n8.44 Chapter 8 HINTS AND EXPLANATIONS \u2234 Vol. of mixture of H 2 and O 2 = 0.015 \u00d7 22400 = 336 ml
Answer: B
\nOriginal PDF solution page
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Solution: n eq of Li OH = Q F \u21d2 w 24 1 2 5 4825 0 8 96500 \u00d7 = \u00d7 \u00d7 . .
Answer: B
\nOriginal PDF solution page
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Solution: n eq Cu deposited at cathode = Q F or, w 63 5 2 12 4 4825 96500 . . \u00d7 = \u00d7 \u21d2 w = 19.685 gm But the increase in mass of cathode is only 19.05 gm It represents that 20 gm of sample contains only 19.05 gm Cu. \u2234 % of Cu = 19 05 20 100 95 25 . . % \u00d7 = Now, Q F n Cu n Fe eq eq = + (oxidised at anode) or, 12 4 4825 96500 19 05 63 5 2 56 2 . . . \u00d7 = \u00d7 + \u00d7 w \u21d2 w = 0.56 gm \u2234 Percentage of Fe = 0 56 20 100 2 8 . . % \u00d7 =
Answer: B
\nOriginal PDF solution page
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Solution: Theoretical n eq of NaOH formed = n eq Cu = 3 18 63 6 2 0 1 . . . \u00d7 = Actual n eq of NaOH formed = \u00d7 = 60 1 1000 0 06 . \u2234 Percentage yeild = 0 06 0 1 100 60 . . % \u00d7 =
Answer: A
\nOriginal PDF solution page
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Solution: Cell reaction during discharge: Pb + PbO 2 + 2H 2 SO 4 2PbSO 4 + 2H 2 O n Pb taken = 200 208 and n PbO taken 2 200 240 = Hence, PbO 2 is L.R. Now, n eq PbO Q F 2 = \u21d2 200 240 2 10 96500 \u00d7 = \u00d7 t \u21d2 t = 16083.33 sec
Answer: A
\nOriginal PDF solution page
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Solution: Q F of I N a S O eq eq 2 2 3 = = \u2212 n n or, i \u00d7 \u00d7 = \u00d7 \u00d7 2 3600 96500 72 1 0 1000 1 . \u21d2 i = 0.965 A
Answer: A
\nOriginal PDF solution page
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Solution: C 14 H 10 + 2H 2 O C 14 H 8 O 2 + 6H + + 6e \u2013 n eq C 14 H 8 O 2 = Q F \u21d2 w 208 6 1 40 60 0 965 96500 \u00d7 = \u00d7 \u00d7 \u00d7 . \u2234 w = 0.832 gm
Answer: B
\nOriginal PDF solution page
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Solution: Number of coulombs required = 1000 0 00033 . per Kg Cu Energy required = \u00d7 = 1000 0 00033 0 33 10 6 . . J \u2234 Cost of electricity = \u00d7 \u00d7 = 4 10 3600 10 1 11 3 6 . Rupee
Answer: D
\nOriginal PDF solution page
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Solution: n eq CH 3 Coo \u2013 oxidised = Q F or, n \u00d7 = \u00d7 \u00d7 \u00d7 1 0 5 482 5 60 0 8 96500 . . . \u21d2 n = 0.12 \u2234 Moles of (C 2 H 6 + CO 2 ) produced = 3 2 0 12 0 18 \u00d7 = . . and total volume = 0.18 \u00d7 22.4 = 4.032 L
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 \u00b0 = \u22c5 = E V 0 059 2 10 0 177 6 . log .
Answer: A
\nOriginal PDF solution page
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Solution: Back EMF = \u22c5 = \u00d7 \u2212 0 06 2 0 12 0 08 5 4 10 3 . log . . . V
Answer: B
\nOriginal PDF solution page
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Solution: Equivalent of charge used = 0 4825 10 3600 96500 0 18 . . \u00d7 \u00d7 = F Cell reaction during charge: Cu Zn Cu Zn + \u23af \u2192 \u23af + + + 2 2 100 1 1000 0 1 \u00d7 = . mole 100 1 1000 0 1 \u00d7 = . mole = 0.2 eq = 0.2 eq Final 0.2 \u2013 0.1 8 = 0.2 + 0.18 = 0.02 eq = 0.38 eq = 0.01 mole = 0.19 mole \u2234 Final [ ] . . Zn M 2 0 01 100 1000 0 1 + = \u00d7 = and [ ] . . Cu M 2 0 19 100 1000 1 9 + + \u00d7 = Now, cell reaction as galvanic cell:\n8.45 Electrochemistry HINTS AND EXPLANATIONS Zn + Cu 2+ Zn 2+ + Cu and E E n Zn Cu cell cell = \u00b0 \u2212 \u22c5 + + 0 06 2 2 . log [ ] [ ] = \u2212 \u2212 \u2212 \u22c5 = [ . ( . )] . log . . . 0 34 0 76 0 06 2 0 1 1 9 1 1084 V
Answer: C
\nOriginal PDF solution page
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Solution: Reactions involved are 2 2 2 2 H O H O 2 electrolysis \u23af \u2192 \u23af\u23af\u23af\u23af + N 2 + 3H 2 2NH 3 NH 3 + 2O 2 HNO 3 + H 2 O NH 3 + HNO 3 NH 4 NO 3 For 1 mole NH 4 NO 3 , 3 moles of H 2 O should be electrolyzed. Hence for 1152 Kg NH 4 NO 3 , moles of H 2 needed = \u00d7 \u00d7 = \u00d7 3 1152 10 80 4 32 10 3 4 . Now, n eq H Q F 2 = \u21d2 4.32 \u00d7 10 4 \u00d7 2 = i \u00d7 \u00d7 24 3600 96500 \u2234 i = 96500 A/day
Answer: A
\nOriginal PDF solution page
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Solution: n eq HC 3 H 5 O 3 = n OH Q F \u2212 = or, w 90 1 50 10 1158 96500 3 \u00d7 = \u00d7 \u00d7 \u2212 \u21d2 w = 0.054 gm \u2234 % of lactice acid = 0 054 1 100 5 4 . . % \u00d7 =
Answer: A
\nOriginal PDF solution page
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Solution: n n n H O NH S O formed 2 2 4 2 2 8 = = ( ) and n eq NH S O Q F ( ) 4 2 2 8 = \u21d2 n i \u00d7 = \u00d7 = \u00d7 \u00d7 2 102 34 2 3600 0 5 96500 . [ ] 2 2 4 2 2 8 2 SO S O e \u2212 \u2212 \u2212 \u23af \u2192 \u23af + \u21d2 i = 321.67 A
Answer: B
\nOriginal PDF solution page
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Solution: n n n As H AsO = = 3 3 and n eq H 3 AsO 3 = n eq I 2 = Q F or, w 75 2 1 68 10 96 5 96500 3 \u00d7 = \u00d7 \u00d7 \u2212 . . \u21d2 w = 6.3 \u00d7 10 \u20135 gm
Answer: D
\nOriginal PDF solution page
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Solution: w E Q F = \u21d2 52 2 87 2 19 3 2 3600 96500 . . \u00d7 = \u00d7 \u00d7 \u00d7 \u03b7 \u21d2 \u03b7 = 0.8333 or 83.33 %
Answer: B
\nOriginal PDF solution page
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Solution: n eq Cu 2+ reduced = Q F \u21d2 n \u00d7 2 = 2 10 19 3 60 96500 3 \u00d7 \u00d7 \u00d7 \u2212 . \u2234 n = 1.2 \u00d7 10 \u20135 \u2234 [ ] ( . ) . CuSO M 4 0 5 5 1 2 10 2 250 1000 9 6 10 = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212
Answer: A
\nOriginal PDF solution page
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Solution: w E Q F = \u21d2 12 3 123 6 . \u00d7 = \u00d7 Q 0.5 F \u21d2 Q = 1.2 F + 6H + + 6e \u2013 + 2H2O NH2 NO2 and Energy consumed = 1.2 F \u00d7 3.0 V = 347.4 KJ
Answer: C
\nOriginal PDF solution page
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Solution: n eq H 2 (at cathode) = n eq O 2 + n eq H 2 S 2 O 8 (at anode) or, 9 08 22 7 2 2 27 22 7 4 194 2 . . . . \u00d7 = \u00d7 + \u00d7 w \u2234 w = 38.8 gm
Answer: C
\nOriginal PDF solution page
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Solution: Initial mass of H 2 SO 4 , w 1 = 3600 \u00d7 1.5 \u00d7 40 100 2160 = gm Final mass of H 2 SO 4 , w 2 = 3600 \u00d7 1.1 \u00d7 10 100 396 = gm \u2234 Moles of H 2 SO 4 consumed = w w 1 2 98 18 \u2212 = Now, n eq H 2 SO 4 = Q F \u21d2 18 \u00d7 1 = ( ) amp-hr \u00d7 3600 96500 \u2234 Number of ampere-hr = 482.5
Answer: B
\nOriginal PDF solution page
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Solution: \u2227 \u2227 = = \u22c5 \u22c5 \u22c5 \u2217 \u2217 m m NaCl KCl NaCl NaCl KCl NaCl KCl /C) /C) R G C R G ( ) ( ) ( ( \u03ba \u03ba 1 1 1 \u22c5 \u22c5 = \u22c5 \u22c5 1 C R C) R C) KCl KCl NaCl ( ( or, \u2227 = \u00d7 \u00d7 m NaCl) ( . . 120 200 0 1 6400 0 003 \u21d2 \u2227 m(NaCl) = 125 \u03a9 \u20131 cm \u20131 mol \u20131 2 2 4 2 2 8 2 SO S O e \u2212 \u2212 \u2212 \u2192 + \u23a1 \u23a3 \u23a4 \u23a6\n8.46 Chapter 8 HINTS AND EXPLANATIONS
Answer: C
\nOriginal PDF solution page
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Solution: \u2227\u00b0 = \u00d7 \u00b0 + + \u2212 m m m NH CrO NH CrO [( ) ] ( ) ( ) 4 2 4 4 4 2 2 \u03bb \u03bb = (2 \u00d7 6.6 \u00d7 10 \u20138 + 5.4 \u00d7 10 \u20138 ) \u00d7 96500 = 0.01795 \u03a9 \u20131 m 2 mol \u20131
Answer: A
\nOriginal PDF solution page
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Solution: \u2227 = \u22c5 \u2227\u00b0 = m m C \u03b1 \u03ba or, 0.9 \u00d7 4.25 \u00d7 10 \u20132 = 382 5 3 . C 10 \u00d7 \u21d2 C = 0.1 M
Answer: C
\nOriginal PDF solution page
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Solution: \u2227 = m K C \u21d2 1.5 \u00d7 10 \u20132 = 3 06 10 2 56 10 3 3 . . \u00d7 \u2212 \u00d7 \u2212 \u2212 C \u2234 C mol m mol l V = = \u00d7 \u22c5 = \u2212 \u2212 \u2212 1 30 1 30 10 585 58 5 3 3 1 / . \u2234 V = 3 \u00d7 10 5 L
Answer: D
\nOriginal PDF solution page
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Solution: \u2227 = = \u22c5 \u2217 m C G G C \u03ba \u21d2 100 1 0 5 1 5 0 1 10 3 = \u00d7 \u00d7 \u2212 R . . . \u21d2 R ohm = 100 3 Now, V = IR \u21d2 I V R A = = = 5 100 3 0 15 / .
Answer: C
\nOriginal PDF solution page
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Solution: Ionic mobility, \u03bc \u03bb \u00b0 = = \u00b0 speed of ion Pot. gradient F m or, speed 19 3 5 50 96500 . \u239b \u239d \u239c \u239e \u23a0 \u239f = \u21d2 speed = 2 \u00d7 10 \u20133 cm/s
Answer: A
\nOriginal PDF solution page
Open page 439 in PDFExtracted text
Solution: \u2227 eq = C \u03ba \u21d2 150 3 4 10 1 6 10 5 6 6 \u00d7 \u00d7 \u2212 \u00d7 \u2212 \u2212 . . \u2234 S = 1.2 \u00d7 10 \u20138 mol cm \u20133 = 1.2 \u00d7 10 \u20135 M
Answer: B
\nOriginal PDF solution page
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Solution: K = G \u22c5 G* \u21d2 \u03ba 1 4 1 280 1 50 . / / = \u21d2 \u03ba = 0.25 s m \u20131 for 0.5 M Now, \u2227 = = \u00d7 = \u00d7 \u2212 \u2212 m C s m mol \u03ba 0 25 0 5 10 5 10 3 4 2 1 . .
Answer: D
\nOriginal PDF solution page
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Solution: Ag A S Ag A ( ) ( )M \u001e \u21c0 \u001e \u21bd \u001e \u001e + + \u2212 + x y x M Ag B S Ag B M ( ) ( )M \u001e \u21c0 \u001e \u21bd \u001e \u001e + + \u2212 + x y y Now, ( x + y ) \u22c5 x = 3 \u00d7 10 \u201314 and ( x + y ) \u22c5 y = 1 \u00d7 10 \u201314 \u2234 [Ag ] , [A ] . ; [B ] . + \u2212 \u2212 \u2212 \u2212 \u2212 = + = \u00d7 = = \u00d7 = = \u00d7 x y x y 2 10 1 5 10 0 5 10 7 7 7 M M M Now, \u03ba solution = \u00d7 = \u00d7 \u00d7 \u00d7 + \u00d7 \u00d7 \u00d7 + \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 3 75 10 2 10 10 60 1 5 10 10 80 0 5 10 8 7 3 7 3 . . . \u2212 \u2212 \u2212 \u00b0 \u00d7 7 3 10 \u03bb B \u2234 \u03bb B \u00b0 \u2212 \u2212 = 135 ohm cm mol 1 2 1
Answer: A
\nOriginal PDF solution page
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Solution: \u2227 \u00b0 = \u2227\u00b0 + \u2227\u00b0 \u2212 \u2227 \u00b0 eq eq eq eq [Be (Po ) ] [BeCl ] [K Po ] [K ] 3 4 2 2 3 4 Cl = + \u2212 = \u2212 \u2212 160 140 100 200 1 2 1 ohm cm eq Now, n eq = \u21d2 = \u00d7 \u2212 \u03ba C C 200 1 2 10 5 . \u21d2 C = 6 \u00d7 10 \u20138 eq cm \u20133 = 6 \u00d7 10 \u20135 N = 10 \u20135 M Now, K sp = = \u00d7 = \u00d7 \u2212 \u2212 108 108 10 1 08 10 5 5 5 23 S ( ) .
Answer: B
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+ "title": "Chem Sec 2",
+ "originalName": "Section B - Multi Correct",
+ "questions": [
+ {
+ "question_id": "electrochemistry-chem-sec-2-1-81",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": null,
+ "subject": "chemistry",
+ "chapter": "electrochemistry",
+ "chapterTitle": "Electrochemistry",
+ "type": "mcqm",
+ "rawChapterType": "MSQ",
+ "originalNumber": 81,
+ "displayNumber": 1,
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+ "identifier": "C",
+ "content": ""
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+ "identifier": "D",
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+ }
+ ],
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+ "A",
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: A, C
\nOriginal PDF solution page
Open page 439 in PDFExtracted text
Solution: Net cell reaction is spontaneous in electrochemical cell but non-spontaneous in electrolytic cell. Cathode is +ve in electrochemical cell but \u2013ve in electrolytic cell.
Answer: A, C, D
\nOriginal PDF solution page
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Solution: For the cell: Ag(s) | Ag Cl (s) | Cl\u2212 || Ag + | Ag(s), the net cell reaction is Ag + + Cl\u2212 \u001f Ag Cl (s).
Answer: C, D
\nOriginal PDF solution page
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Solution: Left electrode: Ag(s) + Cl \u2212 (aq) \u2192 Ag Cl (s) + e \u2212 1 \u00d7 2 Right electrode : Hg 2 Cl 2 (s) + 2e \u2212 \u2192 2Hg(l) + 2 Cl \u2212 (aq) Net reaction: 2Ag(s) + Hg 2 Cl 2 (s) \u2192 2Ag Cl(s) + 2Hg (l)\n8.47 Electrochemistry HINTS AND EXPLANATIONS
Answer: B, C
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Solution: n n n n eq eq eq eq Cu Mg Na Al = \u00d7 = = \u00d7 = = \u00d7 = = 63 5 63 5 2 2 24 24 2 2 11 5 23 1 0 5 . . ; . . ; 9 9 27 3 1 \u00d7 =
Answer: A, C
\nOriginal PDF solution page
Open page 440 in PDFExtracted text
Answer: B, C
\nOriginal PDF solution page
Open page 440 in PDFExtracted text
Solution: \u2227 \u00b0 eq = 60 + 80 = 140 ohm \u20131 cm 2 eq \u20131 \u2227 \u00b0 m = 140 \u00d7 6 = 840 ohm \u20131 cm 2 eq \u20131
Answer: B, C, D
\nOriginal PDF solution page
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Solution: Salt bridge does not change standard potential of any electrode.
Answer: A, B, C
\nOriginal PDF solution page
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Solution: Moles of electron involved = 0 25 9 65 3600 96500 0 09 . . . \u00d7 \u00d7 = \u2234 Mass of Zn involved = 0 09 2 65 4 2 943 . . . \u00d7 = gm Mass of MnO 2 involved = 0.09 \u00d7 87 = 7.83 gm Mass of NH 4 + involved = 0.09 \u00d7 18 = 1.62 gm
Answer: A, C
\nOriginal PDF solution page
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Solution: Net cell reaction is Cd(s) + 2AgCl(s) \u001f 2Ag(s) + Cd 2+ (aq) + 2Cl \u2013 (aq) \u0394 \u00b0 = \u2212 \u00b0 = \u2212 \u00d7 \u00d7 = \u2212 \u00b0 G 50 2 96500 0 6 115800 C nFE J . \u0394 \u00b0 = \u2212 \u00b0 = \u2212 \u00d7 \u00d7 = \u2212 \u00b0 G 0 2 96500 0 7 135100 C nFE J . \u0394 \u00b0 = \u22c5 \u00b0 \u2212 \u00b0 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u2212 = \u2212 S T T nF E E J/K 2 1 2 1 2 96500 0 6 0 7 50 386 . . \u0394 H \u00b0 = \u0394 G \u00b0 + T \u22c5 \u0394 S \u00b0 = (\u2013135100) + 273 \u00d7 (\u2013386) = \u2013240478 J
Answer: B, C
\nOriginal PDF solution page
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Solution: Anode: 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e \u2013 Cathode: 2H 2 O(l) + 2e \u2013 H 2 (g) +2OH \u2013 (aq) n eq H + produced = n eq OH \u2013 produced = Q F or, n n H OH + \u2212 \u00d7 = = \u00d7 \u00d7 = 1 1 25 965 60 96500 0 75 . . Anode: HPO H H PO 4 2 2 4 \u2212 + \u2212 + \u001e \u21c0 \u001e \u21bd \u001e \u001e 1.0 M 0.75 M 1.0 M Final 0.25 M 1.75 M \u2234 P P HPO H PO H K = + = + = \u2212 \u2212 a log [ ] [ ] . log . . . 4 2 0 2 4 2 15 0 25 0 75 1 30 Cathode: H PO OH HPO H O 2 4 4 2 2 \u2212 \u2212 \u2212 + + \u001e \u21c0 \u001e \u21bd \u001e \u001e 1.0 M 0.75 M 1.0 M Final 0.25 M 0 1.75 M \u2234 P P HPO H PO H K = + = + = \u2212 \u2212 a log [ ] [ ] . log . . . 4 2 0 2 4 2 15 1 75 0 25 3 0
Answer: A, B, D
\nOriginal PDF solution page
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Solution: V, Fe and Hg will be oxidised by NO 3 \u2212
Answer: A, B
\nOriginal PDF solution page
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Solution: Resistance and heat capacity depends on quantity.
Answer: A, B
\nOriginal PDF solution page
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Answer: A, B, C, D
\nOriginal PDF solution page
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Solution: Net cell reaction of discharge is Pb + PbO 2 + 2H 2 SO 4 2PbSO 4 + 2H 2 O x mole x mole Initial mass of H 2 SO 4 , w 1 = 1000 \u00d7 1.26 \u00d7 40 100 504 = gm Final mass of H 2 SO 4 , w 2 = (1260 \u2013 98 x + 18 x ) \u00d7 28 100 = (352.8 \u2013 22.4 x ) gm From reaction, 504 \u2013 98 x = 352.8 \u2013 22.4 x \u21d2 x = 2
Comprehension 1 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
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+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
+ "content": ""
+ },
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+ "identifier": "D",
+ "content": ""
+ }
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+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: E\u00b0 Cell = E E H H Zn Zn \u00b0 \u2212 \u00b0 + + / / 2 2 because E H H \u00b0 + / 2 was higher or, 0.76 = 1.00 \u2013 E Zn Zn \u00b0 + 2 / \u21d2 E Zn Zn \u00b0 + 2 / = 0.24 V
Comprehension 1 - subquestion 2
",
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+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 440 in PDFExtracted text
Solution: E\u00b0 Cell = E E Cu Cu H H \u00b0 \u2212 \u00b0 + + 2 2 / / because E Cu Cu \u00b0 + 2 / was higher or, 0.34 = E Cu Cu \u00b0 + 2 / \u2013 1.00 \u21d2 E Cu Cu \u00b0 + 2 / = 1.34 V
Comprehension 1 - subquestion 3
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 440 in PDFExtracted text
Solution: E\u00b0 Cell = E E Cu Cu Zn Zn \u00b0 \u2212 \u00b0 + + 2 2 / / = 1.34 \u2013 0.24 = 1.10 V\n8.48 Chapter 8 HINTS AND EXPLANATIONS Comprehension II
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: E E Ag K Ag Ag Ag Ag sp + + = \u00b0 \u2212 \u22c5 = \u2212 + / / . log [ ] . . .log 0 06 1 1 0 80 0 06 1 1 = + 0.314 V
Comprehension 2 - subquestion 2
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: E E K V I |Ag I|Ag Ag Ag sp \u00b0 = \u00b0 \u2212 \u22c5 = \u2212 \u2212 + / . log . 0 06 1 0 172
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: E E I I |Ag I|Ag I /Ag I Ag \u2212 \u2212 = \u00b0 \u2212 \u22c5 \u2212 / . log[ ] 0 06 1 = \u20130.172 \u2013 0.06 \u22c5 log 0.04 = \u20130.088 V Comprehension III
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: As(0.40) > (\u20130.87), reduction of Ni 2 O 3 (s) will occur.
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: E\u00b0 cell = (0.40) \u2013 (\u20130.87) = 1.27 V
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: Net cell reaction is independent from OH \u2013 (aq)
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: \u2013 \u0394 G \u00b0 = nFE\u00b0 cell = 2 \u00d7 96500 \u00d7 1.27 = 245110 J Comprehension IV
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: H 3 O + is only needed in balancing cathode reaction: NO H O e HNO H O 3 3 2 2 3 2 4 \u2212 + \u2212 + + \u23af \u2192 \u23af +
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: Moles of electron needed = 2 \u00d7 moles of HNO 2 formed
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: n eq HNO Q F 2 = \u21d2 0 1 2 10 96500 . \u00d7 = \u00d7 t \u21d2 t = 1930 sec Comprehension V
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: \u0394 G\u00b0 = \u2013 n FE\u00b0 \u21d2 \u2013237.39 \u00d7 10 3 = \u20132 \u00d7 96500 \u00d7 E\u00b0 cell \u2234 E \u00b0cell = 1.23 V
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: Moles of H 2 needed = 23 739 237 39 0 1 . . . = \u2234 Volume of H 2 needed = 0.1 \u00d7 22.7 = 2.27 L
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: E cell is independent from [OH \u2013 ]
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: \u0394 \u00b0 = \u0394 \u00b0 \u2212 \u0394 \u00b0 = \u2212 \u00d7 \u2212 \u2212 \u00d7 S H G T ( . ) ( . ) 285 8 10 237 39 10 298 3 3 = \u2013162.4 J/K \u20131
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: \u03b7 = \u2212\u0394 \u00b0 \u2212\u0394 \u00b0 = = ( ) ( ) . . . G H 237 39 285 8 0 8306 or 83.06% Comprehension VI
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: Refer theory given is passage.
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: E E E V O H O H Fe Fe \u00b0 = \u00b0 \u2212 \u00b0 = \u2212 \u2212 = + + 2 2 2 1 229 0 447 1 676 / , / . ( . ) .
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: E E Cu Cu Pb Pb \u00b0 > \u00b0 + + 2 2 / / and hence Cu will not oxidise easily.
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: n eq Fe Q F = \u21d2 w 56 2 0 5 1 0 3600 96500 \u00d7 = \u00d7 \u00d7 . . \u21d2 w = 0.522 gm
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: Theoretical Comprehension VII
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: n eq Ni Q F = \u21d2 w 58 7 2 15 3600 0 6 96500 . . \u00d7 = \u00d7 \u00d7 \u21d2 w = 9.85 gm
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 441 in PDFExtracted text
Solution: V = A \u00d7 t \u21d2 9 85 8 9 4 0 2 . . ( . ) = \u00d7 \u00d7 t \u21d2 t = 0.138 cm\n8.49 Electrochemistry HINTS AND EXPLANATIONS
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: n eq H Q F 2 = \u21d2 V H 2 22 4 2 15 3600 0 4 96500 . . \u00d7 = \u00d7 \u00d7 \u21d2 V L H 2 2 5 = .
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: Anode produced is only O 2 gas. n eq O Q F 2 = \u21d2 w 8 15 3600 96500 = \u00d7 \u21d2 w = 4.477 gm Comprehension VIII
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: E\u00b0 cell = 0.8 \u2013 0.05 = 0.75 V Now, E RT nF K cell \u00b0 = \u22c5 ln \u21d2 0 75 1 2 38 92 . . ln = \u00d7 \u22c5 K \u21d2 ln K = 58.38
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: The oxidation reaction of glucose contains H + and E E H = \u00b0 \u2212 \u22c5 + 0 0592 2 2 . log[ ] \u21d2 E \u2013 E\u00b0 = 0.0592 P H = 0.6512 V
Comprehension 10 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: Standard potential is independent from ammonia concentration. Comprehension IX
Comprehension 11 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: Left electron: H 2 (g) 2H + (aq) + 2e \u2013 Right electrode: 2AgCl(s) + 2e \u2013 2Ag(s) + 2Cl \u2013 (aq) \u2234 Net reaction: H 2 (g) + 2AgCl(s) 2Ag(s) + 2H + (aq) + 2Cl \u2013 (aq)
Comprehension 11 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: \u0394 \u00b0 = \u22c5 \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u2212 = \u2212 S nF E E T T J/K 2 1 2 1 2 96500 0 21 0 23 20 193 . . \u0394 G\u00b0 = \u2013nFE\u00b0 = \u20132 \u00d7 96500 \u00d7 0.23 = \u2013 44390 J at 15\u00b0C Now, \u0394 H \u00b0 = \u0394 G \u00b0 + T \u22c5 \u0394 S \u00b0 = (\u201344390) + 288 \u00d7 (\u2013193) = \u201399974 J = \u201349987 J/mole AgCl
Comprehension 11 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: \u0394 S\u00b0 = \u2013193 J/K = \u201396.5 J/K per mole AgCl
Comprehension 12 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: \u0394 G \u00b0 298 = \u0394 H \u00b0 \u2013 T \u22c5 \u0394 S \u00b0 = (\u201349987) \u2013 298 \u00d7 (\u201396.5) or, \u20131 \u00d7 96500 \u00d7 E\u00b0 = \u2013 21230 \u21d2 E\u00b0 = 0.22 V = E Cl AgCl/Ag \u00b0 \u2212 / or, E E K Cl AgCl/Ag Ag Ag sp \u00b0 = \u00b0 + \u2212 + / / . log 0 058 1 or, 0 22 0 80 0 058 1 . . . log = + K sp \u21d2 K sp = 1 \u00d7 10 \u201310 Hence, Solubility, S = K M sp = \u2212 10 5 Comprehension X
Comprehension 12 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: Add Ag(s) in both sides of given reaction to get cell reaction. Now, for E\u00b0 cell \u0394 G \u00b0 = \u2013nFE\u00b0 \u21d2 [\u2013109] \u2013 [77 + (\u2013129)] \u00d7 10 3 = \u20131 \u00d7 96500 \u00d7 E\u00b0 cell \u2234 E\u00b0 cell = 0.59 V
Comprehension 12 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: E K cell eq \u00b0 = 0 059 . log n \u21d2 0 59 0 059 1 1 . . log = \u22c5 K sp \u2234 K sp = 10 \u201310
Comprehension 13 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: Zn(s) + 2Ag + (aq) \u00a1 Zn 2+ (aq) + 3Ag(s); E\u00b0 = 0.80 \u2013 (\u20130.76) Now, E K eq \u00b0 = \u22c5 0 059 . log n \u21d2 1 56 0 059 2 2 2 . . log [ ] [ ] = \u22c5 + + Zn Ag \u2234 log [ ] [ ] . Zn Ag 2 2 52 88 + + =
Comprehension 13 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 442 in PDFExtracted text
Solution: Moles of Zn added = 6 539 10 65 39 10 2 3 . . \u00d7 = \u2212 \u2212 Moles of Ag + present = 10 1000 100 10 5 6 \u2212 \u2212 \u00d7 = (L.R.) \u2234 Moles of Ag precipitated = 10 \u20136\n8.50 Chapter 8 HINTS AND EXPLANATIONS Comprehension XI
Comprehension 13 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: For KCl: \u2227 = = \u22c5 \u2217 m C G G C \u03ba \u21d2 G* = \u2227 m \u22c5 C/G or, G* = \u2227 m \u22c5 C \u22c5 R = 200 \u00d7 (0.02 \u00d7 10 \u20133 ) \u00d7 100 = 0.4 cm \u20131
Comprehension 14 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: \u03ba water G G cm = \u22c5 = \u00d7 = \u00d7 \u03a9 \u2217 \u2212 \u2212 \u2212 1 10000 0 4 4 10 5 1 1 .
Comprehension 14 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: For NaCl: \u2227 = = \u22c5 \u2217 m C G G C \u03ba or, 125 1 8000 1 10000 0 4 585 58 5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f . / . V \u21d2 V = 1.25 \u00d7 10 8 cm 3 = 1.25 \u00d7 10 5 L Comprehension XII
Comprehension 14 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: E E Cl Cl I \u00b0 > \u00b0 \u2212 \u2212 2 2 / /I
Comprehension 15 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: E E Mn Mn O H O, H \u00b0 > \u00b0 + + + 3 2 2 2 / / Comprehension XIII
Comprehension 15 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: Net reaction: M + (1M) M + (0.05 M); Higher Conc. Lower Conc. E cell > 0, \u0394 G cell < 0
Comprehension 15 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: 70 mV = E RT F \u00b0 \u2212 \u22c5 ln . 0 05 1 (1) E E RT F E RT F req = \u00b0 \u2212 \u22c5 = \u00b0 \u2212 \u22c5 ln . ln ( . ) 0 0025 1 0 05 1 2 (2) and E\u00b0 =
Comprehension 16 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Hence, E req = 140 mV Comprehension XIV
Comprehension 16 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: \u0394 G cell = \u2013nFE cell = \u20132 \u00d7 96500 \u00d7 0.059 = \u201311387 J
Comprehension 17 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: E E M M cell cell Left Right = \u00b0 \u2212 \u22c5 + + 0 059 2 2 2 . log [ ] [ ] or, 0 059 0 0 059 2 4 0 001 1 3 . . log ( / ) . / = \u2212 \u22c5 K sp \u21d2 K sp = 4 \u00d7 10 \u201315 Comprehension XV
Comprehension 17 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: H + + Cl \u2013 + (NaOH) Na + + Cl \u2013 + H 2 O Conductance first decreases and H + ions are replaced by Na + ions. After equivalent point, conductance increase due to increase in number of ions.
Comprehension 18 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: CH 3 COOH + (NaOH added) CH 3 COO \u2013 + Na + + H 2 O As number of ions increases, conductance increases. slight decrease initially was due to some dissociated CH 3 COOH. After equivalence point, in place of CH 3 COO \u2013 ion, number of OH \u2013 ions increases and hence slope becomes greater.
Comprehension 18 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: H + + Cl \u2013 + (NH 4 OH added) NH Cl H O 4 2 + \u2212 + + As H + ions are replaced by NH 4 + ions, conductance decreases. After equivalent point, it become almost constant as the dissociation of added NH4OH will be suppressed in presence of NH 4 + ions.
Comprehension 19 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: HCl will neutralize fi rst followed by CH 3 COOH.
Comprehension 19 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: Ionic mobilities of Ag + and K + ions do not differ largely
Answer: A
\nOriginal PDF solution page
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Solution: CuCl 2 Cu + Cl 2
Answer: C
\nOriginal PDF solution page
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Solution: E E informative Zn Zn Ag Ag \u00b0 < \u00b0 + + 2 / / ( )
Answer: B
\nOriginal PDF solution page
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Solution: Reason is different charges on ions.
Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Solution: Negative reduction potential means greater tendency to get oxidised.
Answer: C
\nOriginal PDF solution page
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Solution: Number of ions increases considerably only for weak electrolytes.
Answer: A
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: It is due to very high over voltage potential of hydrogen at mercury cathode.
Answer: C
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: E E RT nF Q = \u2212 \u22c5 \u00b0 log
Answer: B
\nOriginal PDF solution page
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Solution: Cl \u2013 will combine with Ag + to precipitate AgCl.
Answer: B
\nOriginal PDF solution page
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Solution: At Cathode: 2H O(l) + 2e H (g) + 2OH aq 2 2 \u2212 \u2212 \u23af \u2192 \u23af ( ) At Anode: 2 2 2 3 2 6 2 CH COO aq C H CO e \u2212 \u2212 ( ) ( ) ( ) \u23af \u2192 \u23af + + g g
Answer: C
\nOriginal PDF solution page
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Answer: A \u2192 P, Q; B \u2192 P, Q; C \u2192 Q, R; D \u2192 P, S
\nOriginal PDF solution page
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Answer: A \u2192 R; B \u2192 S; C \u2192 Q; D \u2192 P
\nOriginal PDF solution page
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Answer: A \u2192 R; B \u2192 S; C \u2192 P; D \u2192 Q
\nOriginal PDF solution page
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Solution: E E E V Fe Fe Fe Fe Fe Fe 3 3 2 2 1 2 1 2 0 037 + + + + \u00b0 \u00b0 \u00b0 = \u00d7 + \u00d7 + = \u2212 | , | . E V H ( H O OH ) . . . 4 4 4 2 0 40 1 23 0 83 \u2192 + \u00b0 + \u2212 = \u2212 = \u2212 E E E Cu Cu Cu Cu (Cu Cu ) | Cu | . . . . 2 2 2 0 34 0 52 2 1 0 52 0 + + + + + + \u2192 \u00b0 \u00b0 \u00b0 = \u2212 = \u2212 \u2212 \u2212 = \u2212 7 70 V E V Cr Cr 3 2 3 0 74 2 0 91 3 2 0 4 + + \u00b0 = \u00d7 \u2212 \u2212 \u00d7 \u2212 \u2212 = \u2212 , ( . ) ( . ) .
Answer: A \u2192 R, S; B \u2192 P, R; C \u2192 Q, S
\nOriginal PDF solution page
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Solution: For concentration cell, both half cell must have same configuration. P. Ag Cl AgCl sponteneous, K K eq sp + \u2212 + \u23af \u2192 \u23af = >> 1 1 Q. Ag Br + Cl AgCl Br Cl Non-Spontene eq sp sp \u2212 \u2212 \u23af \u2192 \u23af + = << ; (Ag Br) (Ag ) , K K K 1 o ous R. Ag Ag + + \u23af \u2192 \u23af ( . M) ( . M) 1 0 0 1 Higher to lower concentration, spontaneous S. Cl Cl \u2212 \u2212 \u23af \u2192 \u23af \u2212 ( . M) ( . M) Non spontaneous 0 1 1 0
Answer: 4
\nOriginal PDF solution page
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Solution: E = E P H \u00b0 \u22c5 + \u2192 + + \u2212 \u2212 0 06 2 2 2 2 2 . log [H ] (assuming H e H ) n \u2212 = \u2212 \u22c5 \u21d2 = + + \u2212 0 18 0 0 06 2 1 10 2 3 . . log [H ] [H ] M C H NH H O C H NH H O M 6 5 3 2 6 5 2 3 + \u2212 + + + ( )M C x x \u001e \u21c0 \u001e \u21bd \u001e \u001e h x c = = = \u2212 10 0 04 4 3 1 40 . % or\n8.52 Chapter 8 HINTS AND EXPLANATIONS
Answer: 5
\nOriginal PDF solution page
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Solution: E E = \u2212 \u22c5 \u00b0 + 0 06 2 1 . log [Cu ] 0 31 0 34 0 06 2 1 0 1 2 2 . . . log [Cu ] [Cu ] . M = \u2212 \u22c5 \u21d2 = + + \u2234 [OH ] . \u2212 + \u2212 \u2212 = = = \u21d2 = K sp H Cu p 2 19 9 10 0 1 10 5
Answer: 5
\nOriginal PDF solution page
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Solution: E E E Fe Fe Fe Fe Fe Fe 3 2 3 2 3 2 3 2 3 0 04 2 0 44 + + + + \u00b0 \u00b0 \u00b0 = \u00d7 \u2212 \u00d7 \u2212 = \u00d7 \u2212 \u2212 \u00d7 \u2212 | | | ( . ) ( . ) 1 1 0 76 = . V Now, E E Fe Fe Fe Fe 3 2 3 2 0 06 1 2 3 + + + + = \u2212 \u00b0 + + | | . log [Fe ] [Fe ] or, 0 7 8 0 76 0 06 5 2 3 2 3 . . . log [Fe ] [Fe ] [Fe ] [Fe ] 1 = \u2212 \u21d2 = + + + +
Answer: 5
\nOriginal PDF solution page
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Solution: 2Fe 3+ + 2I \u2013 \u2192 2Fe 2+ + I 2 ; 0.5 M excess 100 % 0 1.0 M 0.5 M Equ. CM 1.0 M 0.5 M E cell V; \u00b0 = \u2212 = 0 77 0 53 0 24 . . . K eq = 10 8 Now, 10 0 5 1 0 5 10 8 2 2 2 5 = \u00d7 \u21d2 = \u00d7 \u2212 ( . ) ( . ) C C M
Answer: 4
\nOriginal PDF solution page
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Solution: Assuming the cell as concentration cell, net cell reaction is Ag C Ag C sp sp(AgI) + \u2212 + = \u239b \u239d \u239c \u239e \u23a0 \u239f \u23af \u2192 \u23af = C K K 1 2 (AgCl) [ l ] ( ) and E C C cell O = \u2212 0 06 1 2 1 . log or, 0.102 = \u2013 0.06 \u22c5 log . . [Cl ] [Cl ] 8 1 10 1 8 10 4 10 17 10 4 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 M
Answer: 1
\nOriginal PDF solution page
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Solution: n n eq eq Pb Tl = + Q F \u21d2 \u00d7 \u00d7 + \u00d7 \u00d7 = \u00d7 5 0 70 100 208 2 5 0 30 100 204 1 1 1 96500 . . . t \u2234 t = 3597.4 sec ; 1 hr
Answer: 0
\nOriginal PDF solution page
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Solution: n Q F x x eq r I = \u21d2 \u00d7 = \u00d7 \u00d7 \u21d2 = 0 36 192 0 075 2 3600 96500 3 . . Now, x + 6(\u20131) = y \u21d2 y = \u20133
Answer: 9
\nOriginal PDF solution page
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Solution: 2 2 2 2 2 NaCl + 2H O NaOH H Cl electrolysis \u23af \u2192 \u23af\u23af\u23af\u23af + + 2 2 2 NaOH + Cl NaCl NaClO H O \u23af \u2192 \u23af + + n NaClO formed = n Cl 2 produced from electrolysis or, ( . ) . . . 10 10 1 0 7 45 100 74 5 2 2 5 96500 3 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 t ( n -factor of Cl 2 in electrolysis) \u2234 t = 7.72 \u00d7 10 5 sec = 8.93 days \u2248 9 days
Answer: 3
\nOriginal PDF solution page
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Solution: Cathode: Cu 2+ + 2e \u2013 Cu Anode 2H 2 O O 2 + 4H + + 4e \u2013 Moles of e \u2013 used = Q F = \u00d7 \u00d7 \u00d7 \u2212 0 161 5 60 96500 5 10 4 . \u001a Eq. of Cu 2+ present = 500 0 1 1000 2 0 1 \u00d7 \u00d7 = . . ( ) excess \u2234 Moles of H + produced = Moles of e \u2013 = 5 \u00d7 10 \u20134 \u2234 [H + ] fi nal = 5 10 500 1000 10 4 3 \u00d7 \u00d7 = \u2212 \u2212 M \u21d2 P H = 3.0
Answer: 2
\nOriginal PDF solution page
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Solution: cathode: Cu 2+ + 2e \u2013 Cu 250 0 1 1000 \u00d7 . 5 1351 96500 \u00d7 = 0.025 mole = 0.07 mole As Cu + will not remain fi nally in solution, no complex formation.
Answer: 3
\nOriginal PDF solution page
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Solution: n eq metal = n eq Cl 2 \u21d2 52 8 9 08 22 7 2 . . . E = \u00d7 \u21d2 E = 66 At. wt. (approx) = 6 4 0 032 200 . . = \u2234 valency At wt Eq.wt = = . . 200 66 3 \u001a\n8.53 Electrochemistry HINTS AND EXPLANATIONS
Answer: 2
\nOriginal PDF solution page
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Solution: Initial mass of H 2 SO 4 , w 1 = 2000 1 1 16 100 352 \u00d7 \u00d7 = . gm Final mass of H 2 SO 4 , w 2 = 2000 1 42 40 100 1136 \u00d7 \u00d7= . gm Now, n eq H 2 SO 4 produced = Q F or, ( ) 1136 352 98 1 965 9 3600 96500 \u2212 \u00d7 = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f i \u21d2 i = 2A
Answer: 9
\nOriginal PDF solution page
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Solution: Ionic mobility, \u03bc \u03bb = \u00b0 = m F Speed of ion Potential gradient or, 7 5 10 96500 1 93 0 12 3 . / . / . \u00d7 = \u00d7 \u2212 distance 20 3600 \u21d2 distance = 0.09 m = 9 cm
Answer: 8
\nOriginal PDF solution page
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Solution: n eq Ag oxidised = Q F \u21d2 w 108 1 9 65 1 3600 96500 \u00d7 = \u00d7 \u00d7 . \u2234 w = 38.88 gm \u2234 Mass of anode dissolved = 38 88 100 60 64 8 . . \u00d7 \u00d7 gm \u2234 Final mass of anode = 72.8 \u2013 64.8 = 8 gm
Answer: 9
\nOriginal PDF solution page
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Solution: Cathode: 2H 2 O(l) + 2e \u2013 H 2 (g) + 2OH \u2013 (aq) Anode: 2ce \u2013 (aq) Cl 2 (g) + 2e \u2013 n eq OH\u2013 produced = Q F \u21d2 n OH \u2212 \u00d7 = \u00d7 = \u2212 1 9 65 10 96500 10 3 . \u2234 [OH \u2013 ] fi nal = 10 100 10 3 5 \u2212 \u2212 = M \u21d2 P H = 9.0 Four digit integer type
Answer: 0287
\nOriginal PDF solution page
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Solution: \u039b\u00b0 = \u00b0 + \u00b0 \u2212 m m m (AgCl) (AgCl) (Cl ) \u03bb \u03bb = \u00d7 + \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 6 19 10 7 81 10 14 00 10 3 3 3 1 2 1 . . . ohm m mol Now, \u039b m C S = \u21d2 \u00d7 = \u00d7 \u2212 \u2212 \u03ba 14 10 2 8 10 3 4 . \u21d2 S = 0.02 mol m \u20133 = 2 \u00d7 10 \u20135 M = 287 \u00d7 10 \u20135 g/L
Answer: 0142
\nOriginal PDF solution page
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Solution: [S ] K [H S] [ ] . ( ) 2 1 2 2 2 8 13 3 16 10 10 0 1 10 10 \u2212 + \u2212 \u2212 \u2212 \u2212 = \u22c5 \u22c5 = \u00d7 \u00d7 = a a K H M \u2234 [Ag ] [S ] + \u2212 \u2212 \u2212 \u2212 = = \u00d7 = \u00d7 K M sp 2 48 16 16 4 10 10 2 10 Now, E Ag /Ag + = \u2212 \u00d7 \u2212 0 80 0 06 1 1 2 10 16 . . log = \u20130.142 V \u2234 Required potential = 142 mV
Answer: 0815
\nOriginal PDF solution page
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Solution: 2Hg(l) + 2Fe 3+ \u001f Hg Fe 2 2 2 2 + + + excess 10 \u20133 M Equ. 10 \u20133 \u2013 x x 2 x = \u00d7 \u2212 10 100 10 3 M \u2234 x = 9 \u00d7 10 \u20134 Now, K eq = \u00d7 \u2212 = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 x x x 2 10 9 10 2 9 10 1 10 9 10 2 2 3 2 4 4 2 4 2 3 4 ( ) ( ) ( ) Now, E E E K cell Fe /Fe Hg /Hg eq 3+ 2 2+ \u00b0 = \u00b0 \u2212 \u00b0 = 0 06 2 . log or, 0 7724 0 06 2 9 10 2 3 4 . . log \u2212 \u00b0 = \u22c5 \u00d7 \u2212 E Hg /Hg 2 2+ \u2234 E V Hg /Hg 2 2+ \u00b0 = 0 815 .
Answer: 0571
\nOriginal PDF solution page
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Solution: The cell reaction is 6 14 6 2 7 3 3 Fe Cr O H Fe Cr H O 2+ 2 7 2 2 + + \u2192 + + \u2212 + + + E E n cell cell = \u00b0 \u2212 \u22c5 + + + + + 0 06 3 6 3 2 2 6 2 7 2 8 . log [Fe ] [Cr ] [Fe ] [Cr O ][H ] = \u2212 \u2212 \u00d7 \u00d7 \u00d7 ( . . ) . log ( . ) ( ) ( . ) ( ) 1 35 0 77 0 06 6 0 75 4 0 75 2 1 6 2 6 8 = 0.571 V
Answer: 0090
\nOriginal PDF solution page
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Solution: Cell reaction: H 2 (g) + 2Ag + \u2192 2H + + 2Ag(s) E E P cell cell H 2 = \u00b0 \u2212 \u22c5 + + 0 06 2 2 2 . log [H ] [Ag ]\n8.54 Chapter 8 HINTS AND EXPLANATIONS or, 0.50 = 0.80 \u2013 0.03 log 1 1 2 2 \u00d7 + [Ag ] \u21d2 [Ag + ] = 10 \u20135 M \u2234 Mass of Ag in alloy = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = \u00d7 \u2212 \u2212 250 10 1000 108 2 7 10 5 4 . gm \u2234 % of Pb in alloy 2 7 10 2 7 10 2 7 10 100 90 3 4 3 . . . % \u00d7 \u2212 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u2212
Answer: 0100
\nOriginal PDF solution page
Open page 447 in PDFExtracted text
Solution: E K cell eq \u00b0 = 0 06 . log n \u21d2 (0.2 \u2013 0.08) = 0 06 1 . log K eq \u21d2 K eq = 100
Answer: 2633
\nOriginal PDF solution page
Open page 447 in PDFExtracted text
Solution: Left electrode: Mn(s) \u2192 Mn 2+ + 2e \u2013 Right electrode: 2H 2 O(l) \u2192 O 2 (g) + 4H + + 4e \u2013 \u2234 Net cell reaction: 2Mn (s) + 2H 2 O(l) \u2192 2Mn e+ + O 2 (g) + 4H + Now, E E Mn P cell cell 2+ O = \u00b0 \u2212 \u22c5 \u22c5 + 0 06 4 1 2 4 2 . log [ ] [H ] = \u2212 \u2212 \u2212 \u00d7 \u00d7 [ . ( . )] . log ( . ) ( . ) . 1 229 1 185 0 06 4 0 001 0 01 0 25 1 2 4 = 2.633 V
Answer: 0193
\nOriginal PDF solution page
Open page 447 in PDFExtracted text
Solution: \u0394 S nF E T P = \u22c5 \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 = 2 96500 0 001 193 . J/k
Answer: 1300
\nOriginal PDF solution page
Open page 447 in PDFExtracted text
Solution: \u0394 \u03a3 \u0394 \u03a3 \u0394 r f f G nFE G G \u00b0 = \u2212 \u00b0 = \u00b0 \u2212 \u00b0 Products Reactants or, \u2212 \u00d7 \u00d7 = \u00d7 \u00b0 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 \u00d7 + \u00d7 + \u00d7 \u2212 + \u00d7 \u2212 12 9600 2 5 1000 4 4 0 3 0 6 280 4 4 . [ ( ) (OH) \u0394 f G Al ( ( . )] \u2212 156 25 \u2234 \u0394 f G \u00b0 = \u2212 \u2212 Al KJ/mol (OH) 4 1300
Answer: 0028
\nOriginal PDF solution page
Open page 447 in PDFExtracted text
Solution: n eq MnO 2 = Q F \u21d2 8 7 87 1 3 99 10 96500 3 . . \u00d7 = \u00d7 \u00d7 \u2212 t \u2234 t = 2.418 \u00d7 10 6 sec \u2248 28 days
Answer: 1520
\nOriginal PDF solution page
Open page 447 in PDFExtracted text
Solution: SnCl 2 Sn 2+ + 2Cl \u2013 Cathode: Sn 2+ + 2e \u2013 Sn Anode: 2Cl \u2013 Cl 2 + 2e\u2013 Cl 2 + SnCl 2 SnCl 4 Moles of SnCl 2 taken = 19 190 0 1 = . Moles of Sn produced = 1 19 119 0 01 . . = = moles of Cl 2 produced = moles of SnCl 4 formed and moles of SnCl 4 left = 0.1 \u2013 (0.01 + 0.01) = 0.08 \u2234 m m SnCl SnCl 2 4 0 08 190 0 01 261 1520 261 = \u00d7 \u00d7 = . .
Answer: 0772
\nOriginal PDF solution page
Open page 447 in PDFExtracted text
Solution: n eq Au = Q F \u21d2 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 80 8 0 10 19 7 3 197 2 4 96 500 4 . . . t \u2234 t = 772 sec
Answer: 0055
\nOriginal PDF solution page
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Solution: AgBr(s) \u001f Ag + + Br \u2013 (10 \u20137 + x )M x M Now, (10 \u20137 + x ) x = 12 \u00d7 10 \u201314 \u21d2 x = 3 \u00d7 10 \u20137 Final solution: [Ag + ] = 4 \u00d7 10 \u20137 M, [Br \u2013 ] = 3 \u00d7 10 \u20137 M; [ ] NO M 3 7 10 \u2212 \u2212 = Now, \u03ba solution = \u03bb \u00b0 m (Ag + ) \u00d7 [Ag + ] + \u03bb \u00b0 m (Br \u2013 ) \u00d7 [Br \u2013 ] + \u03bb \u00b0m ( ) [ ] NO NO 3 3 \u2212 \u2212 \u00d7 = 6 \u00d7 10 \u20133 \u00d7 (4 \u00d7 10 \u20137 \u00d7 10 3 ) + 8 \u00d7 10 \u20133 \u00d7 (3 \u00d7 10 \u20137 \u00d7 10 3 ) + 7 \u00d7 10 \u20133 (10 \u20137 \u00d7 10 \u20133 )
Answer: 0728
\nOriginal PDF solution page
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Solution: C 60 + 60O 2 60 CO 2 Moles of O 2 needed = 60 60 96 60 12 8 60 \u00d7 = \u00d7 \u00d7 = n C n eq O 2 = n eq azobenzene \u21d2 8 \u00d7 4 = w 182 8 \u00d7 \u21d2 w = 728 gm + 8H + + 8e \u2013 + 4H2O N N NO2 2
Answer: 0108
\nOriginal PDF solution page
Open page 447 in PDFExtracted text
Solution: \u039b\u00b0 eq[Ba (PO ) ] 3 4 2 = 160 + 140 \u2013 100 = 200 Ohm \u20131 cm 2 eq \u20131 Now, \u2227 \u00b0 eq = \u2227 eq = \u03ba C \u21d2 200 1 2 10 5 5 = \u00d7 \u2212 . \u2234 S = 6 \u00d7 10 \u20138 eq/cm 3 = 6 \u00d7 10 \u20135 N = 10 \u20135 M \u2234 K sp = 108 S 5 = 10 8 \u00d7 10 \u201325 M 5
![\"Gaseous](\"./assets/chapter-packs/gaseous-state/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 167 in PDFExtracted text
Solution: P gas + 15.6 = 53.3 + 76.3 \u21d2 P gas = 114 cm Hg = 1.5 atm
Answer: D
\nOriginal PDF solution page
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Solution: Fractional increase = V V V 2 1 1 \u2212 = V V 2 1 1 \u2212 = P P 1 2 1 \u2212 = H H H + \u2212 7 7 1 = 1 7
Answer: D
\nOriginal PDF solution page
Open page 167 in PDFExtracted text
Solution: P 2 P 1 44.5 46 P 0 P 0 45.25 45.25 P 0 \u00d7 45.25 = P 1 \u00d7 46 = P 2 \u00d7 44.5 P 1 + 5 sin30\u00b0 = P 2 P 0 45 25 46 \u00d7 . + 5 \u00d7 1 2 = P 0 45 25 44 5 \u00d7 . . \u21d2 P 0 = 75.4 cm Hg
Answer: C
\nOriginal PDF solution page
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Solution: P 1 V 1 = P 2 V 2 \u21d2 10 \u00d7 2A = (10 + h ) \u00d7 h A \u21d2 h = 1.71 m
Answer: C
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: dv dt = V 0 273 = 0.08 \u21d2 V 0 = 21.84 L
Answer: A
\nOriginal PDF solution page
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Solution: V T 1 1 = V T 2 2 \u21d2 1 0 0 . x + = 0 6 100 . ( ) x + \u2212 \u21d2 x = 250 \u21d2 0 K = \u2013250\u00b0C
Answer: C
\nOriginal PDF solution page
Open page 167 in PDFExtracted text
Solution: V T 1 1 = V T 2 2 \u21d2 V T = V V T T + + \u0394 \u0394 \u21d2 \u0394 \u0394 V V T . = 1 T \u21d2 y = 1 x
Answer: B
\nOriginal PDF solution page
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Solution: V T 1 1 = V T 2 2 \u21d2 V t 1 1 273 + = 1.1 V t 1 2 273 + \u2234 Percentage increase in temperature = t t t 2 1 1 \u2212 \u00d7 100 = ( ) 10 2730 1 + t %
Answer: D
\nOriginal PDF solution page
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Solution: Number of SO 2 molecules = N \u21d2 Number of atoms = 3N
Answer: C
\nOriginal PDF solution page
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Solution: n n N O 2 2 = V V N O 2 2 \u21d2 m m N O 2 2 28 32 = 1 7 8 \u21d2 m m N O 2 2 = 1 1
Answer: A
\nOriginal PDF solution page
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Solution: V n 1 1 = V n 2 2 \u21d2 4 3 10 2 8 3 \u03c0 ( ) = 4 3 2 1 3 \u03c0 d ( ) \u21d2 d = 5 cm
Answer: A
\nOriginal PDF solution page
Open page 167 in PDFExtracted text
Solution: P = P CO 2 + P air = 0 5 0 0821 300 1 1 . . \u00d7 \u00d7 + = 13.315 atm
Answer: D
\nOriginal PDF solution page
Open page 167 in PDFExtracted text
Solution: Weight of fi lled balloon, W = 20 g + 40 \u00d7 0.6 = 44 g Weight of displaced air, B = 40 \u00d7 1.3 = 52 g \u2234 Balloon will lift upward with pay load = 52 \u2013 44 = 8 g B W\n3.45 Gaseous State HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: Water will behave like ideal gas on disappearance of intermolecular forces. V = nRT P = 4 5 10 18 3 . \u00d7 \u00d7 (22.4 \u00d7 10 \u20133 ) m 3 = 5.6 m 3
Answer: C
\nOriginal PDF solution page
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Solution: d = m v \u21d2 1.5 = n n n n co co co co \u00d7 + \u00d7 + \u00d7 \u00d7 28 44 0 0821 300 1 2 2 ( ) . \u21d2 n co = 7 055 8 945 . . n co 2 Alkali will absorb all CO 2 . Hence, final pressure is due to CO. P co = n n n co co co + 2 \u00d7 P total = 7 055 7 055 8 945 760 . . . + \u00d7 mm = 335.1 mm
Answer: C
\nOriginal PDF solution page
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Solution: V V water vapour water = 1 0 0821 373 1 18 0 96 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f . . l ml = 1633.24
Answer: C
\nOriginal PDF solution page
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Solution: V O 2 = 3 2 32 0 0821 310 1 . . \u00d7 \u00d7 = 2.5451 L V CO 2 = 8 8 44 0 0821 310 1 . . \u00d7 \u00d7 = 5.0902 L
Answer: B
\nOriginal PDF solution page
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Solution: n CO 2 = 200 0 1 1000 \u00d7 . = 0.02 \u2234 V CO 2 = 0.02 \u00d7 22.4 = 0.448 L
Answer: A
\nOriginal PDF solution page
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Solution: d d d ( ) \u03c1 = M RT = 1.2 \u00d7 10 \u20135 Kg m \u20133 Pa \u20131 \u21d2 M 8 314 300 . \u00d7 = 1.2 \u00d7 10 \u20135 \u2234 M air = 0.03 Kg/mol = 30 gm/mol Now, 30 = n n n n N O N O 2 2 28 + 32 + \u00d7 \u00d7 2 2 \u21d2 n n N O 2 2 : = 1:1
Answer: D
\nOriginal PDF solution page
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Solution: m T P 1 1 1 = m T P 2 2 2 \u21d2 4 \u00d7 T P = m T P 2 2 \u00d7 2 \u21d2 m 2 = 16 gm Hence, (16 \u2013 4) = 12 gm gas should be added.
Answer: B
\nOriginal PDF solution page
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Solution: 74 5 50 . = 1.49 times
Answer: C
\nOriginal PDF solution page
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Solution: V 1 d 1 = V 2 d 2 \u21d2 1500 \u00d7 1.25 = 3.92 \u00d7 d 2 \u21d2 d 2 = 478.3 kg/mol
Answer: A
\nOriginal PDF solution page
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Solution: P V T 1 1 1 = P V T 2 2 2 \u21d2 P r T \u00d7 4 3 1 3 \u03c0 = P r T 4 4 3 2 3 \u00d7 \u03c0 2 \u21d2 r 2 = 2 r 1 \u2234 % Increase in radius = r r r 2 1 1 \u2212 \u00d7 100 = 100 %
Answer: B
\nOriginal PDF solution page
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Solution: Constant = P 2 V = nRT V V \uf8eb \uf8ed \uf8ec \uf8f6 \uf8f8 \uf8f7 2 \u21d2 T V 2 = Constant \u2234 On expansion, temperature will increase.
Answer: C
\nOriginal PDF solution page
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Solution: r R M = \u21d2 r r r n H N e 2 > > 2
Answer: B
\nOriginal PDF solution page
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Solution: P \u00d7 3 = 7 28 0 0821 300 \u00d7 \u00d7 . \u21d2 2.0525 atm
Answer: B
\nOriginal PDF solution page
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Solution: P 10 atm 2 atm 4 L 20 L 1 V 2 T 4 L 20 L 12 L V 1 2 40 R T 2 atm 10 atm 6 atm P 40 R For T max , V = 12 L and P = 6 atm and hence, T max = 6 12 1 0 08 \u00d7 \u00d7 . = 900 K.\n3.46 Chapter 3 HINTS AND EXPLANATIONS
Answer: A
\nOriginal PDF solution page
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Solution: 2Al + 2NaOH + 2H 2 O \u2192 2NaAlO 2 + 3H 2 2 mole 3 mole \u2234 0.15 27 mole 3 2 \u00d7 0.15 27 mole \u2234 V H 2 = 1 5 0 15 27 0 0831 300 0 831 . . . . \u00d7 \u00d7 \u00d7 = 0.25L
Answer: A
\nOriginal PDF solution page
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Solution: N 2 \u2192 2N Initial mole a = 1 4 28 . 0 Final mole a \u2013 0.4 a 2 \u00d7 0.4 a = 0.6 a = 0.8 a Final total moles = 0.6 a + 0.8 a = 1.4 \u00d7 1 4 28 . = 0.07 \u2234 P = 0 07 0 0821 1800 5 . . \u00d7 \u00d7 \u001e 2.07 atm
Answer: B
\nOriginal PDF solution page
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Solution: CH 4 (g) + 2O 2 (g) 127 \u00b0 \u23af \u2192 \u23af\u23af C CO 2 (g) + 2H 2 O (g) As there is no change in mole of gases, P T 1 1 = P T 2 2 \u21d2 1 300 = P 2 400 \u21d2 P 2 = 1.33 atm
Answer: C
\nOriginal PDF solution page
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Solution: P c = X c P total \u21d2 10 \u2013 (1 + 3) = n C 10 10 \u00d7 \u21d2 n c = 6 \u2234 Mass of C = 6 \u00d7 2 = 12 gm
Answer: A
\nOriginal PDF solution page
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Solution: x x M R 4 + 5 = 760 2.4 300 \u2212 \u00d7 \u00d7 and x R 4 = 19 2.4 15 \u00d7 \u00d7 \u2234 M = 96
Answer: B
\nOriginal PDF solution page
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Solution: C 2 H 6 + 7 2 O 2 2CO 2 + 3H 2 O( l ) 1 vol 7 2 vol 2 vol 0 vol \u2234 10 ml 35 ml 20 ml 0 Final volume should be 20 + (40 \u2013 35) = 25 ml but it is 26 ml. Hence, volume occupied by water vapour is (26 \u2013 25) = 1 ml. \u2234 Vapour pressure of water = 1 26 \u00d7 1 atm = 760 26 = 29.23 mm Hg
Answer: C
\nOriginal PDF solution page
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Solution: n H 2 O vapour needed = ( . ) . 26 463 24 1 760 0 0821 300 \u2212 \u00d7 \u00d7 \u00d7 = 1.32 \u00d7 10 \u20134
Answer: B
\nOriginal PDF solution page
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Solution: P = 1 2 0 0821 300 18 50 760 . . \u00d7 \u00d7 \u00d7 \u00d7 = 24.96 mm Hg
Answer: B
\nOriginal PDF solution page
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Solution: Mass of water lost per day = \u2206 P.V RT M \u00d7 = ( ) . 45 5 760 0 0821 310 18 \u2212 \u00d7 \u00d7 \u00d710000 = 372.23 gm
Answer: C
\nOriginal PDF solution page
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Solution: Vapour pressure is a function of temperature only
Answer: D
\nOriginal PDF solution page
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Solution: After achievement of equilibrium with its liquid form which will form on continuous injection of vapour, the pressure due to vapours become constant.
Answer: B
\nOriginal PDF solution page
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Solution: Rate of evaporation will remain constant throughout because neither surface area nor temperature are changing
Answer: A
\nOriginal PDF solution page
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Solution: r r x y = 1 5 and r r y z = 1 6 \u21d2 r r z x = 30 1
Answer: D
\nOriginal PDF solution page
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Solution: Smaller the rate of diff usion of HX, more closer to the HX end, NH 4 X will form.
Answer: C
\nOriginal PDF solution page
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Solution: r r N H 2 2 = M M H N 2 2 \u21d2 \u0394 \u0394 P P t / / 60 = 2 28 \u21d2 t = 16.04 min
Answer: B
\nOriginal PDF solution page
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Solution: M dry air > M moist air
Answer: A
\nOriginal PDF solution page
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Solution: r r CH HBr 4 = P P CH HBr 4 M M HBr CH 4 \u21d2 1 1 = n n CH HBr 4 81 16 \u21d2 n n CH HBr 4 = 0.4 \u2234 X CH 4 = n n n CH CH HBr 4 4 + = 0.31
Answer: C
\nOriginal PDF solution page
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Solution: As HCl will diff use slowly, white fumes will form closer to HCl end.
Answer: B
\nOriginal PDF solution page
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Solution: In gases, the intermolecular distance is much higher than the size of molecules.
Answer: A
\nOriginal PDF solution page
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Solution: u u av,2 av,1 = T T 2 1 = 375 250 = 1.22
Answer: C
\nOriginal PDF solution page
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Solution: u u rms, O rms, O 2 = 3R 2T 16 3RT 32 \u00d7 2 1 \u21d2 u rms, o = 2 V\n3.47 Gaseous State HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: Average speed for a gas depends on temperature and it is independent from the presence of other gas.
Answer: A
\nOriginal PDF solution page
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Solution: Difference in any two kind of speed, \u2206 u K T = \u00d7 Now, d u dT ( ) \u0394 = K 2 T \u21d2 On increasing temperature, \u0394 u decreases.
Answer: D
\nOriginal PDF solution page
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Solution: u u av x av y , , = 2 1 = T T X Y \u21d2 T T X Y = 4 1 Now, P P X Y = nRT V nRT V X X Y Y = T T V V X Y Y X \u00d7 = 4 1 2 1 \u00d7 = 8 1
Answer: A
\nOriginal PDF solution page
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Solution: 1 \u00d7 V = 1 M R T A \u00d7 \u00d7 \u21d2 M M B A = 4 1 0.5 \u00d7 V = 2 M RT B \u00d7 \u2234 u u av A av B , , = M M B A = 2 1
Answer: B
\nOriginal PDF solution page
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Solution: u av, A = u av, B \u21d2 8 RT M A A \u03c0 = 3 RT M B \u21d2 M M B A = 3 8 \u03c0 Now, u av, A = u av, B \u21d2 8 RT M A A \u03c0 = 8 RT M B B \u03c0 \u21d2 T T A B = M M A B = 8 3 1 \u03c0 <
Answer: C
\nOriginal PDF solution page
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Solution: 1 4 . N *. u av = 1 4 6 10 22 4 10 8 8 314 273 28 10 23 3 3 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 . . \u03c0 = 3.05 \u00d7 10 27 m \u20132 s \u20131
Answer: D
\nOriginal PDF solution page
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Solution: u u O O rms rms , , 2 3 = 3 600 32 48 3 300 R R \u00d7 \u00d7 \u00d7 = 3 \u21d2 u rms,O 2 = 3 v m/s
Answer: B
\nOriginal PDF solution page
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Solution: Average translational K.E. per gm = 3 2 RT M
Answer: D
\nOriginal PDF solution page
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Solution: T M A A = T M B B \u21d2 u rms = 3 RT M = Same for both
Answer: C
\nOriginal PDF solution page
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Solution: 3 2 KT = qV \u21d2 3 2 8 314 6 022 10 23 \u00d7 \u00d7 \u00d7 . . T = 1.602 \u00d7 10 \u201319 \u00d7 3 \u21d2 T = 23207.2 K
Answer: B
\nOriginal PDF solution page
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Solution: u 2 rms \u2260 u 2 av
Answer: B
\nOriginal PDF solution page
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Solution: Z w = 1 4 . N *. u av = 1 4 8 \u00d7 \u00d7 P N RT RT M A . \u03c0 \u21d2 Z w \u221d 1 T
Answer: A
\nOriginal PDF solution page
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Solution: u u rms, CH rms, SO 4 2 = 3 16 64 3 300 R T R \u00d7 \u00d7 \u00d7 = 4 1 \u21d2 T = 1200 K \u2234 Average K.E. per mole = 3 2 RT = 3 2 2 1200 \u00d7 \u00d7 = 3600 cal
Answer: A
\nOriginal PDF solution page
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Solution: u av \u221d T \u21d2 u u 2 1 = 432 300 = 1.2
Answer: C
\nOriginal PDF solution page
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Solution: Mole of gas cannot change.
Answer: C
\nOriginal PDF solution page
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Solution: Collision number, Z 1 = 2 \u03c0\u03c3 2 . u av . N * = 2 \u03c0\u03c3 2 . 8 RT M PN RT A \u03c0 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2234 Z 1 \u221d 1 T Collision frequency, Z 11 = 1 2 \u03c0\u03c3 2 u av . N* 2 = 1 2 \u03c0\u03c3 2 . 8 2 RT M PN RT A \u03c0 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2234 Z 11 \u221d 1 3 2 T Mean free path, \u03bb = 1 2 2 \u03c0\u03c3 N * = RT PN A 2 2 \u03c0\u03c3 . \u21d2 \u03bb \u221d T
Answer: A
\nOriginal PDF solution page
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Solution: \u03bb = RT PN A 2 2 \u03c0\u03c3 \u00d7 = 8 314 300 2 1 5 10 4 1 10 1 013 10 6 022 10 10 2 14 5 23 . ( . ) ( . . ) ( . \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u03c0 ) ) = 1.0 \u00d7 10 7 m
Answer: B
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: Velocity is a vector quality.
Answer: A
\nOriginal PDF solution page
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Solution: dN N = 4 2 3 2 2 2 2 \u03c0 \u03c0 m KT u e du mu KT \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 \u2212 = 2 1 3 2 \u03c0 KT E e dE E KT \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 \u2212\n3.48 Chapter 3 HINTS AND EXPLANATIONS For most probable K.E., d dN N dE ( ) = 0 \u21d2 E = 1 2 KT
Answer: D
\nOriginal PDF solution page
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Solution: Deviation from ideal behavior is maximum at low temperature and high pressure.
Answer: A
\nOriginal PDF solution page
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Solution: Z > 1 for H 2 at 0\u00b0C at all pressure.
Answer: D
\nOriginal PDF solution page
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Solution: Z < 1 at low pressure and Z > 1 at high pressure.
Answer: C
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Solution: V 1 = V \u2013 nb \u21d2 b = V V n \u2212 1 \u21d2 4 6 3 \u00d7 \u00d7 \u03c0 d N A = V V n \u2212 1 \u2234 d = 3 2 1 1 3 ( ) V V nN A \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u03c0
Answer: B
\nOriginal PDF solution page
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Solution: P i = P an V + 2 2 \u21d2 P = P an V i + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f 2 2 Greater the value of \u2018 a \u2019, smaller will be \u2018 P \u2019.
Answer: D
\nOriginal PDF solution page
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Solution: Gaseous mixture is always homogeneous.
Answer: A
\nOriginal PDF solution page
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Solution: P 1 V 1 = P 2 V 2 \u21d2 0.5 \u00d7 2000 = 100 \u00d7 V 2 \u21d2 V 2 = 10 ml < 13 ml As the real volume is greater than ideal, the volume occupied by the molecule is significant.
Answer: A
\nOriginal PDF solution page
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Solution: When attractive forces are dominant, V real < V ideal .
Answer: A
\nOriginal PDF solution page
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Solution: B = b a RT \u2212 = 0.03 \u2013 1 344 0 0821 273 . . \u00d7 = \u2013 0.03 l/mol
Answer: A
\nOriginal PDF solution page
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Solution: Z = PV RT m = V V b m m \u2212 = 10 10 b b b \u2212 = 10 9
Answer: A
\nOriginal PDF solution page
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Solution: P an V V nb + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u2212 2 2 ( ) = nRT may be expressed as P a d M M d b + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f . 2 2 = RT as d = m v = n m v \u00d7 Now, P + \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 3 6 2 2 44 44 2 2 0 05 2 2 . ( . ) ( ) . . = 0.0821 \u00d7 300 \u21d2 P = 1.226 atm
Answer: A
\nOriginal PDF solution page
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Solution: B = b a RT \u2212 = \u20131.0 L/mol Now, PV m = RT 1 + \u239b \u239d \u239c \u239e \u23a0 \u239f B V m and d = M V m Hence, PM d = RT 1 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f B d M or, 1 40 \u00d7 d = 0.08 \u00d7 262.5 1 1 0 40 + \u2212 \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ( . ) d \u2234 d = 2.005 g/L
Answer: A
\nOriginal PDF solution page
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Solution: When P \u2192 0, V \u2192 \u221e and hence e a/VRT \u2192 1 and ( V \u2013 b ) \u2192 V . Hence, P = RT V = 0 0821 300 410 5 . . \u00d7 = 0.06 atm
Answer: A
\nOriginal PDF solution page
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Solution: For a van der Waals gas, Z = V V b a V RT m m m \u2212 \u2212 or, 0.8 = 0 5 0 5 0 04 0 5 0 08 300 . . . . . \u2212 \u2212 \u00d7 \u00d7 a \u21d2 a = 3.44 atm L 2 /mol 2
Answer: B
\nOriginal PDF solution page
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Solution: At Boyle\u2019s temperature, dz dp = 0 \u21d2 T = 168 0 35 . = 480 K
Answer: B
\nOriginal PDF solution page
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Solution: Theory based. The initial slope of Z vs. P curve increases with increase in temperature, above Boyle\u2019s temperature, only upto 2 \u00d7 T B . Then, the slope starts decreasing.
Answer: A
\nOriginal PDF solution page
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Solution: For van der Waals gas, Z = V V b a V RT m m m \u2212 \u2212 \u00d7 At Boyle\u2019s temperature, Z = V V b a V R a Rb m m m \u2212 \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1 2 + \u2212 b V V b m m ( )
Answer: D
\nOriginal PDF solution page
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Solution: Ideal gas can never be liquified.
Answer: A
\nOriginal PDF solution page
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Solution: For ideal behavior, Boyle\u2019s temperature should be closer to 600 K.
Answer: C
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Solution: T P c c = 8 27 27 2 a Rb a b = 8 b R \u2234 T P T P c c c c \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f CO CH 2 4 = b b CO CH 2 4 = 304 72 190 45 = 1 1\n3.49 Gaseous State HINTS AND EXPLANATIONS
Answer: D
\nOriginal PDF solution page
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Solution: At T > T c , the gas can never be liquified.
Answer: D
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: P c V c = 3 8 RT c \u21d2 V c = 3 8 0 0821 128 41 05 \u00d7 \u00d7 . . = 0.096 L/mol
Answer: C, D
\nOriginal PDF solution page
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Solution: Boyle\u2019s law constant = PV = nRT
Answer: B, C
\nOriginal PDF solution page
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Solution: P V RT P V RT 0 0 0 0 0 0 + = PV R T PV RT 0 0 0 0 2 \u00d7 + \u21d2 P = 4 3 P 0 and n = 4 3 2 0 0 0 P V R T \u00d7 \u00d7 = 2 3 0 0 0 P V RT
Answer: A, B, C, D
\nOriginal PDF solution page
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Solution: P \u0192 = P i V V V n + \u239b \u239d \u239c \u239e \u23a0 \u239f \u0394 (a) P \u0192 = 24.2 \u00d7 10 10 1 1 + \u239b \u239d \u239c \u239e \u23a0 \u239f = 22 atm (b) P \u0192 = 24.2 \u00d7 10 10 1 2 + \u239b \u239d \u239c \u239e \u23a0 \u239f = 20 atm (c) P \u03b7 = P 10 10 1 + \u239b \u239d \u239c \u239e \u23a0 \u239f n \u21d2 n = ln ln . \u03b7 1 1 (d) P \u0192 = 24.2 \u00d7 10 10 1 + \u239b \u239d \u239c \u239e \u23a0 \u239f n atm
Answer: B, C
\nOriginal PDF solution page
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Solution: As the average molar mass increases, the molar mass of vapours must be greater than that of N 2 .
Answer: B, C
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Solution: (a) Number of molecules colliding at the wall per unit time per unit area, Z w = 1 4 . u av . N * N * is same for both but u av , He > u av , Ne (b) Average force per collision \u221d Change in momentum \u221d M
Answer: B, C
\nOriginal PDF solution page
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Solution: At valve \u2013 I: P 1 V 1 = P 2 V 2 \u21d2 1 \u00d7 60 A = P 2 \u00d7 45 A \u21d2 P 2 = 1.33 atm < 1.5 atm Hence, valve \u2013 I will not open. At valve \u2013 II: P 1 V 1 = P 2 V 2 \u21d2 1 \u00d7 60 A = P 2 \u00d7 30 A \u21d2 P 2 = 2 atm < 2.2 atm Hence, valve \u2013 II will not open. At valve \u2013 III: P 1 V 1 = P 2 V 2 \u21d2 1 \u00d7 60 A = P 2 \u00d7 20 A \u21d2 P 2 = 3 atm > 2.5 atm Hence, valve \u2013 III will open fi rst. As the piston will reach at valve \u2013 III, the gas will come out till the pressure of gas becomes 2.5 atm. Now, 2 5 20 821 1000 . \u00d7 \u00d7 = n \u00d7 0.0821 \u00d7 300 \u21d2 Moles of gas remained, n = 5 3 At valve \u2013 IV: P 1 V 1 = P 2 V 2 \u21d2 2.5 \u00d7 20 A = P 2 \u00d7 15 A \u21d2 P 2 = 3.33 atm < 4.4 atm Hence, valve - IV will not open. At valve \u2013 V: P 1 V 1 = P 2 V 2 \u21d2 2.5 \u00d7 20 A = P 2 \u00d7 10 A \u21d2 P 2 = 5 atm > 4.8 atm Hence, valve \u2013 V will open until the gas pressure becomes 4.8 atm.
Answer: C
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Solution: u u rms, A rms, B = 3 300 3 400 R M M R A B \u00d7 \u00d7 : = 3 2 \u21d2 M A = M B
Answer: C, D
\nOriginal PDF solution page
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Solution: On increasing the temperature at constant volume, the average speed of molecules as well as number of molecular collisions at wall increases.
Answer: A, D
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Answer: A, B
\nOriginal PDF solution page
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Solution: At very high pressure, P a V + \u239b \u239d \u239c \u239e \u23a0 \u239f 2 \u001e P \u21d2 Z = 1+ b P RT . and Z = PV RT \u21d2 P RT = Z V \u21d2 Z = V V b \u2212\n3.50 Chapter 3 HINTS AND EXPLANATIONS
Answer: A, B, C, D
\nOriginal PDF solution page
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Solution: T c = 273 + (\u2013177) = 96 K \u21d2 T B = 27 8 \u00d7 96 = 324K = 51\u00b0C (a) Z = PV RT m . = 0 821 9 6 0 0821 96 . . . \u00d7 \u00d7 = 1 But at T = T c , Z < 1 at low pressure (b) Z = PV RT m . = 0 821 40 0 0821 400 . . \u00d7 \u00d7 = 1 But at T > T B , Z > 1 at all pressure (c) Z = PV RT m . = 82 1 0 310 0 0821 324 . . . \u00d7 \u00d7 = 0.96 < 1 But at T = T B and P > 50 atm, Z > 1 (d) Z = PV RT m . = 0 821 32 4 10 0 0821 324 3 . . . \u00d7 \u00d7 \u00d7 \u2212 = 10 \u20133 < 1 But at T = T c and P < 50 atm, Z = 1
Answer: A, B, D
\nOriginal PDF solution page
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Solution: a. 9 8 27 36 8 R a Rb \u00d7 \u00d7 = a b. 3 \u00d7 a b 27 2 \u00d7 (3 b ) 2 = a c. 3 8 27 36 8 27 2 \u00d7 \u00d7 a b a Rb \u2260 a d. 27 64 8 27 27 2 2 2 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f R a Rb a b = a
Answer: A, D
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: Real gas may behave ideally at Boyle\u2019s temperature. P = RT V m = R a Rb V m \u00d7 = a bVm .
Answer: A, C
\nOriginal PDF solution page
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Solution: Z = PV RT m = 1 + B \u2032 .P + C \u2032 .P 2 + \u2026. (1) Z = PV RT m = 1 + B Vm + C V m 2 + \u2026. (2) Or, P = RT V m (1 + B V m + C V m 2 + \u2026.) Substituting this value in Equation (1), we get: Z = 1 + B \u2032 . RT V B V C V m m m 1 2 + + + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa ... + C \u2032 . RT V B V C V m m m 1 2 2 + + + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa ... + \u2026 = 1 + \u2032 B RT V m + \u2032 + \u2032 B RT B C RT V m . ( ) 2 2 + \u2026. Comparing this with Equation (2), we get: B \u2032 RT = B and B \u2032 RT . B + C .( RT ) 2 = C
Answer: A, B, C, D
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Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Number of strokes = ( ) ( ) ( ) ( ) 8 bar cm 1 bar cm \u00d7 \u00d7 \u00d7 1000 25 4 3 3 = 80
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: F = P.A = 8 10 5 2 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f N m \u00d7 (4 \u00d7 10 \u20134 m 2 ) = 320 N
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: m = F g = 320 10 = 32 kg\n3.51 Gaseous State HINTS AND EXPLANATIONS Comprehension \u2013 II
Comprehension 1 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: PV m = RT (Let 0K = \u2013 x \u00b0N) 28 = R (0 + x ) x = 233.33 40 = R (100 + x ) \u2234 0K = \u2013233.33\u00b0N
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: R = 28 x = 0.12 L \u2013 atm/K-mol
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: V = nRT P = 2 0 12 66 67 233 33 2 \u00d7 \u00d7 + . ( . . ) = 36 L Comprehension \u2013 III
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: A 20 A B D C 20 60 A B D C B 20 A B D C 20 60 P 1 1 atm 30 A B D C 20 50 P 2 1 atm For AB column: P 1 = 1 atm = 7 6 cm Hg V 1 = 20 A cm 3 P 2 = 1 atm \u2013 20 cm Hg = 76 \u2013 20 = 56 cm Hg V 2 = 30 A cm 3 Now, P 1 V 1 \u2260 P 2 V 2 Hence, only end A is not closed. C 20 A B D C 20 60 P 1 1 atm 30 A B D C 20 50 P 2 1 atm For CD column: P 1 = 1 atm = 76 cm Hg V 1 = 60 A cm 3 P 2 = 1 atm + 20 cm Hg = 76 + 20 = 96 cm Hg V 2 = 50 A cm 3 As P 1 V 1 \u2260 P 2 V 2 , only end D is not closed. Hence, both the ends are closed.
Comprehension 2 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: P 0 P 0 20 20 60 A B D C A B D C P 1 P 2 30 20 50 For AB column: P 0 \u00d7 20 = P 1 \u00d7 30 For CD column: P 0 \u00d7 60 = P 2 \u00d7 50 As P 1 + 20 cm Hg = P 2 or, 20 30 0 P + 20 cm Hg = 60 50 0 P \u21d2 P 0 = 37.5 cm Hg
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: D D C A B P 4 P 3 (80 \u2013 x ) 20 x P 4 + 20 cm Hg = P 3 or, 60 80 0 P x ( ) \u2212 + 20 = 20 0 P x \u21d2 x = 13.88 If both ends are open, then mercury will fall down.\n3.52 Chapter 3 HINTS AND EXPLANATIONS Comprehension \u2013 IV
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Total moles of product gases = PV RT = 410 5 2 9 0 0821 2000 . . . \u00d7 \u00d7 = 7.25 \u2234 Moles of gases per 0.04 mole of nitroglycerine = 0.04 \u00d7 7.25 = 0.29
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Moles of gases except A = 4 75 0 821 0 0821 250 . . . \u00d7 \u00d7 = 0.19 \u2018A\u2019 must be H 2 O because it solidifies at \u201323\u00b0C and its mole = 0.29 \u2013 0.19 = 0.10.
Comprehension 3 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: Moles of gases C and D = 2 1 0 821 0 0821 300 . . . \u00d7 \u00d7 = 0.07 \u2234 Mole of gas \u2018B\u2019, which is CO 2 = 0.19 \u2013 0.07 = 0.12
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: The gas remained, D must be N 2 and its mole = 1 8 0 821 0 0821 300 . . . \u00d7 \u00d7 = 0.06 and gas \u2018C\u2019 is O 2 and its mole = 0.07 \u2013 0.06 = 0.01. Comprehension \u2013 V
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: All H 2 O(g) will solidify in bulb \u2018B\u2019.
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: n H O 2 + n CO 2 + n N 2 = 570 1 642 760 0 0821 300 \u00d7 \u00d7 \u00d7 . . = 0.05 n CO 2 + n N 2 = 0 21 1 642 0 0821 300 0 21 1 642 0 0821 200 . . . . . . \u00d7 \u00d7 + \u00d7 \u00d7 = 0.035 \u2234 n H O 2 = 0.05 \u2013 0.035 = 0.015
Comprehension 4 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: H 2 O(g) will solidify in \u2018B\u2019 as well as \u2018C\u2019 but CO 2 (g) will solidify only in \u2018C\u2019.
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: n N 2 = 22 8 1 642 760 0 0821 1 300 1 200 1 80 . . . \u00d7 \u00d7 + + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = 0.0125 \u2234 n CO 2 = 0.035 \u2013 0.0125 = 0.0225 Comprehension VI
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: Let x mole, NH 4 Cl was present initially. NH 4 Cl(s) \u2192 NH 3 (g) + HCl(g) x mole x mole Now, PV = nRT 114 \u00d7 V = 0.01 \u00d7 R \u00d7 300 and 908 \u00d7 V = (0.01 + 2 x ) \u00d7 R \u00d7 600 \u2234 x \u2248 0.015 \u2234 Mass of NH 4 Cl = x \u00d7 53.5 \u2248 0.8 gm
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: P NH 3 = 908 114 2 2 \u2212 \u00d7 340 mm Hg
Comprehension 5 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: V = 0 01 0 0821 300 760 114 . . \u00d7 \u00d7 \u00d7 = 1.642 L Comprehension \u2013 VII
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: Water will vaporize till P H O 2 = 0.04 atm Now, PV = nRT \u21d2 0.04 \u00d7 (40 \u00d7 10 3 ) = w 18 \u00d7 0.08 \u00d7 300 or w = 1200 = 1.2 kg \u2234 Percentage of water vaporized = 1 2 5 . \u00d7 100 = 24 %
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: V = nRT P = 5000 18 0 08 300 0 04 \u00d7 \u00d7 . . = 1.67 \u00d7 10 5 L Comprehension \u2013 VIII
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: \u2212 dP dt = K ( P \u2013 P 0 ) \u21d2 \u2212 \u2212 \u222b dP P P P P 0 1 2 = K dt t 0 \u222b \u21d2 ln P P P P 1 0 2 0 \u2212 \u2212 = Kt or, ln 20 1 1 2 \u2212 \u2212 P = 0.001 \u00d7 3600 = ln38 \u21d2 P 2 = 1.5 atm
Comprehension 6 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: Number of balloons = ( . ) 20 1 5 10 1 2 \u2212 \u00d7 \u00d7 = 92.5 \u2248 92
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 175 in PDFExtracted text
Solution: ln P P P P 1 0 2 0 \u2212 \u2212 = Kt \u21d2 ln 20 1 2 1 \u2212 \u2212 = 0.001 \u00d7 t \u21d2 t = 2900 sec\n3.53 Gaseous State HINTS AND EXPLANATIONS Comprehension \u2013 IX
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 176 in PDFExtracted text
Solution: \u2212 dP dt = K.P \u21d2 \u2212 \u222b dP P P 1 5 . atm = K dt t 0 \u222b \u21d2 P = (1.5 atm). e \u2013 Kt
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 176 in PDFExtracted text
Solution: A 38 cm 2 A x x 2 The pressure of gas in closed arm, P = 1 atm + 38 3 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f x cm = 114 3 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f x cm Hg Now, 114 3 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f x = 114 \u00d7 e \u2013 kt \u21d2 x = 76 (1\u2013 e \u2013 kt ) cm Hg
Comprehension 7 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 176 in PDFExtracted text
Solution: x 2 = 38(1 \u2013 e \u2013 kt ) cm Hg Comprehension \u2013 X
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 176 in PDFExtracted text
Solution: u mp = 2 RT M \u21d2 T = M u R \u00d7 mp 2 2 = ( ) ( ) 32 10 400 2 8 3 2 \u00d7 \u00d7 \u00d7 \u2212 = 320K = 47\u00b0C
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 176 in PDFExtracted text
Solution: u rms \u2013 u mp = 400 m/s \u21d2 3 2 RT M RT M \u2212 = 400 m/s or, T = 400 3 2 2 \u2212 \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u00d7 M R = ( ) 400 3 2 2 6 2 10 8 2 3 + \u2212 \u00d7 \u00d7 \u2212 = 400 K = 127\u00b0C
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 176 in PDFExtracted text
Solution: c 1 f ( c ) c 2 C 4 \u03c0 M RT 2 3 2 \u03c0 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 C 1 2 \u00d7 e MC RT \u2212 1 2 2 = 4 \u03c0 M RT 2 3 2 \u03c0 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 C 2 2 \u00d7 e MC RT \u2212 2 2 2 or, C C 1 2 2 2 = e M C C RT ( ) 1 2 2 2 2 \u2212 or, M C C RT ( ) 1 2 2 2 2 \u2212 = 2 ln C C 1 2 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2234 T = M C C R C C ( ) .ln 1 2 2 2 1 2 4 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = ( )( ) ln 28 10 300 600 4 8 300 600 3 2 2 \u00d7 \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = 337.5 K = 64.5\u00b0C
Comprehension 8 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 176 in PDFExtracted text
Solution: c = ? f ( c ) T n . T C 4 2 3 2 2 2 2 \u03c0 \u03c0 M RT c e MC RT \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u2212 = 4 2 3 2 2 2 2 \u03c0 \u03c0 M RT n c e MC RT n . . \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u2212 or, n 3/2 = e MC RT n 2 2 1 1 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, 3 2 ln n = MC RT n n 2 2 1 \u00d7 \u2212 \u2234 C = 3 1 nRT n M n ln ( ) \u2212\n3.54 Chapter 3 HINTS AND EXPLANATIONS
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: f ( c ) T 1 T 2 > T 1 C
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: f ( c ) M 1 M 2 > M 1 C
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: ( ) ( ) dN dN 1 2 = 4 2 2 4 2 3 2 2 2 2 3 2 \u03c0 \u03c0 \u03c0 \u03c0 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 M RT u e du N M RT M u RT ( ) ( ) mp mp 2 2 2 2 2 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u00d7 u e du N M u RT mp mp = 4 \u00d7 e Mu RT mp \u2212 \u2212 2 2 4 1 ( ) =
Comprehension 9 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: e \u20133 Comprehension \u2013 XI
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: n A = m 2 , n B = m 16 , n C = m 32
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: Z W = 1 4 \u00d7 u av \u00d7 N * = 1 4 8 RT M \u03c0 \u00d7 N * \u21d2 Z W \u221d N M *
Comprehension 10 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: \u03bb = 1 2 2 \u03c0\u03c3 N * \u21d2 \u03bb A : \u03bb B : \u03bb C = 1 1 2 1 2 16 1 2 32 2 2 2 \u00d7 \u00d7 \u00d7 m m m : : = 1 : 2 : 4
Comprehension 11 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: E total = 3 2 nRT and n max for A
Comprehension 11 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: Z = 1 for all (Ideal behaviour)
Comprehension 11 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: Z 1 = 2 \u03c0 \u03c3 2 u av \u00d7 N * Z A : Z B : Z C = 1 2 \u00d7 1 2 2 \u00d7 m : 2 2 \u00d7 1 16 16 \u00d7 m : 2 2 \u00d7 1 32 32 \u00d7 m = 1 2 2 1 64 1 128 2 : :
Comprehension 12 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 177 in PDFExtracted text
Solution: u av \u221d 1 M Comprehension \u2013 XII For critical point, dP dV = 0 and d P dV 2 2 Now, dP dV = 0 \u21d2 \u2212 \u2212 + RT V b a T V ( ) . 2 3 2 = 0 \u21d2 RT V b a T V ( ) . \u2212 = 2 3 2 (1) and d P dV 2 2 = 0 \u21d2 2 6 3 4 RT V b a T V ( ) . \u2212 \u2212 = 0 \u21d2 2 3 3 4 RT V b a T V ( ) . \u2212 = (2) From (1) \u00f7 (2) : V \u2013 b = 2 3 V \u21d2 V C = 3 b Eq 1 : RT b b ( ) 3 2 \u2212 = 2 3 3 a T b .( ) \u21d2 T C = 8 27 a Rb\n3.55 Gaseous State HINTS AND EXPLANATIONS and P C = RT V b a T V \u2212 \u2212 . 2 = aR b 216 3
Comprehension 12 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: T C = 8 27 a Rb
Comprehension 12 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: P C = aR b 216 3
Answer: A
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: Avogadro\u2019s hypothesis is valid only for gases due to large intermolecular distance.
Answer: A
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: Charle\u2019s law
Answer: C
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: d = PM RT but M is independent from d , P or T .
Answer: D
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: H 2 and Cl 2 are reactive gases.
Answer: C
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: Escaping tendency increases only on increasing the energy of molecules.
Answer: C
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: Graham\u2019s law is valid for ideal as well as non-ideal gases.
Answer: B
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Answer: B
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 178 in PDFExtracted text
Solution: Volume of ideal gas should be the total volume minus the volume occupied by gas molecules.
Answer: A
\nOriginal PDF solution page
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Solution: As the average K.E. is same, increase in mass decreases their speed.
Answer: A
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Solution: Total K.E. = 3 2 nRT As the pressure exerted by the vapour is same in both but volume is in 1 : 2 ratio, the moles is also in 1 : 2 ratio.
Answer: D
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Solution: f ( c ) T 1 T 2 > T 1 C
Answer: A
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Answer: D
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Solution: Excluded volume is \u2018nb\u2019.
Answer: D
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Solution: T C < T B and hence, attractive forces are dominant.
Answer: D
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Solution: PV is constant at constant temperature.
Answer: A
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Solution: Above Boyle\u2019s temperature, gases show positive deviation.
Answer: B
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Answer: B
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Answer: A
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Solution: K.E. of molecules is the function of T.
Answer: A \u2192 P, R; B \u2192 Q ; C \u2192 S ; D \u2192 T
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Solution: Boyle\u2019s law : PV = K \u21d2 dP dV T \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 K V 2 = \u2212 P V And d PV dP T ( ) \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = 0 Charle\u2019s law V T = K \u21d2 dV dT P \u239b \u239d \u239c \u239e \u23a0 \u239f = K = V T Avogadro\u2019s law : V n = K = RT P \u21d2 dV dn P T \u239b \u239d \u239c \u239e \u23a0 \u239f , = RT P Graham\u2019s law r = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f dP dt \u221d 1 d
Answer: A \u2192 Q, R; B \u2192 S; C \u2192 P; D \u2192 Q
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Solution: Average translational K.E. per mole = 3 2 RT Average translational K.E. per gram = 3 2 RT M
Answer: A \u2192 R; B \u2192 P, S; C \u2192 Q
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Solution: T C = 8 27 a Rb a b X \u239b \u239d \u239c \u239e \u23a0 \u239f = 120, a b Y \u239b \u239d \u239c \u239e \u23a0 \u239f = 333.33, a b Z \u239b \u239d \u239c \u239e \u23a0 \u239f = 171.4 V C = 3 b P C = a b 27 2 , a b X 2 \u239b \u239d \u239c \u239e \u23a0 \u239f = 4800, a b Y 2 \u239b \u239d \u239c \u239e \u23a0 \u239f = 11111.11, a b Z 2 \u239b \u239d \u239c \u239e \u23a0 \u239f = 4898
Answer: A \u2192 S; B \u2192 R; C \u2192 Q; D \u2192 P
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Solution: A. P V = P nRT P \u239b \u239d \u239c \u239e \u23a0 \u239f = P nRT 2 P V P\n3.56 Chapter 3 HINTS AND EXPLANATIONS B. P V = nRT V V / = nRT V 2 P V V C. V P = nRT P 2 1 P 2 V P D. P V = P nRT 2 = 10 2 log P nRT P V log P
Answer: A \u2192 Q; B \u2192 S; C \u2192 P; D \u2192 R
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Solution: Moles of water vapour formed, n = PV RT = 22 8 827 6 760 0 0821 300 . ( ) . \u00d7 \u2212 \u00d7 \u00d7 = 1
Answer: A \u2192 Q; B \u2192 R; C \u2192 P; D \u2192 S
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Solution: Pressure correction = a n V . 2 2 = 4 5 10 2 2 \u00d7 = 1 atm Ideal volume = V \u2013 nb = 10 \u2013 5 \u00d7 0.05 = 9.75 L Volume occupied by molecules = nb 4 = 0 25 4 . = 0.0625 L Volume correction = nb = 0.25L
Answer: A \u2192 P, R, S; B \u2192 P, Q, S; C \u2192 P, Q, S; D \u2192 P, R, S
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Answer: A \u2192 P, S; B \u2192 Q, R; C \u2192 Q
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Solution: (A) \u03bb = 1 2 2 \u03c0\u03c3 N * = RT PN A 2 2 \u03c0\u03c3 At constant volume, N * = constant \u21d2 \u03bb \u03bb 2 1 = 1 At constant pressure, \u03bb \u221d T \u21d2 \u03bb \u03bb 2 1 = 2 (B) Z 1 = 2 \u03c0\u03c3 2 \u00d7 u av \u00d7 N * = 2 \u03c0\u03c3 2 \u00d7 8 RT M PN RT A \u03c0 \u239b \u239d \u239c \u239e \u23a0 \u239f At constant volume : Z 1 \u221d T \u21d2 Z Z 1 2 1 1 , , = 2 At constant pressure : Z 1 \u221d 1 T \u21d2 Z Z 1 2 1 1 , , = 1 2 (C) Z 11 = 1 2 2 2 \u03c0\u03c3 . . * u N av = 1 2 8 2 2 \u03c0\u03c3 \u03c0 RT M PN RT A \u239b \u239d \u239c \u239e \u23a0 \u239f At constant volume : Z 11 \u221d T \u21d2 Z Z 11 2 11 1 , , = 2 At constant pressure : Z 11 \u221d 1 3 2 ( ) T \u21d2 Z Z 11 2 11 1 , , = 1 2 2
Answer: A \u2192 P, S; B \u2192 Q; C \u2192 R
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Solution: A. P = 1 atm + 38 cm Hg = 1.5 atm Q. P = 1 atm + 57 cm Hg = 1.75 atm R. P = 38 cm Hg = 0.5 atm S. P = 1 atm + 1.9 m glycerine = 1+ 190 2 72 13 6 76 \u00d7 \u00d7 . . = 1.5 atm
Answer: A \u2192 Q; B \u2192 S; C \u2192 P; D \u2192 R; E \u2192 T
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Solution: T C = 273 + (\u2013177) = 96 K = \u2013177\u00b0C T B = 27 8 96 \u00d7 = 324 K = 51\u00b0C A. V i = 0 2 0 08 96 20 10 3 . . \u00d7 \u00d7 \u00d7 = 76.8 ml But V real < V ideal in given condition \u21d2 V r < 76.8 ml B. Z = 1 \u21d2 V r = V i = 0 2 0 08 324 6 48 10 3 . . . \u00d7 \u00d7 \u00d7 = 800 ml C. Above Boyle\u2019s temperature, Z > 1 \u2234 V r > V i = 0 2 0 08 350 7 10 3 . . \u00d7 \u00d7 \u00d7 = 800 ml D. Below Boyle\u2019s temperature, Z < 1 \u2234 V r < V i = 0 2 0 08 300 6 10 3 . . \u00d7 \u00d7 \u00d7 = 800 ml E. T = T B but P > 50 atm \u21d2 Z > 1 \u2234 V r > V i = 0 2 0 08 324 64 8 10 3 . . . \u00d7 \u00d7 \u00d7 = 80 ml
Answer: 7
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Solution: V T 1 1 = V T 2 2 \u21d2 45 300 = 42 2 T \u21d2 T 2 = 280 K = 7\u00b0C\n3.57 Gaseous State HINTS AND EXPLANATIONS
Answer: 5
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Solution: 1 2 P 1 = 1 atm + hm water = (10 + h ) m water V 1 = 4 3 \u03c0 (1 mm) 3 A = \u03c0 r 2 = \u03c0 mm 2 \u2234 r = 1 mm P 2 = 1 atm = 10 m water V 2 = 2 \u03c0 mm 3 Now, P 1 V 1 = P 2 V 2 \u21d2 (10 + h ) \u00d7 4 3 \u03c0 = 10 \u00d7 2 \u03c0 \u21d2 h = 5 m Hence, water holding capacity of pool, V = 1 3 \u03c0 r 2 h = 1 3 \u03c0 (10 m) 2 \u00d7 5 m = 500 3 \u03c0 m 3
Answer: 5
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Solution: Number of cosmic events = Number of Ar-atoms = 1 911 10 22 7 6 . . \u00d7 \u2212 l l \u00d7 6 \u00d7 10 23 = 5.05 \u00d7 10 16
Answer: 4
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Solution: For lifting of balloon, B > W or, ( V \u00d7 \u03c1 outside air \u00d7 g ) > ( V \u00d7 \u03c1 inside air + m additional ) g or, V ( \u03c1 outside air \u2013 \u03c1 inside air ) > m additional or, 91 1 29 0 08314 290 1 29 0 08314 10 8 314 \u00d7 \u00d7 \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f > . . . T \u2234 T > 293.22 Hence, diff erence in temperature, \u0394 T > 3.22 \u2248 4 K.
Answer: 3
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Solution: n released gas = n taken \u2013 n remained or, m 22 4 1 25 . . \u00d7 = P p \u00d7 \u00d7 \u2212 \u2212 \u00d7 \u00d7 3 0 0821 273 0 8 3 0 0821 273 . ( . ) . \u21d2 m = 3 gm
Answer: 3
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Solution: I T Vacuum VL 4 atm 300 K VL II I VL 600 K P atm VL 600 K ( P + 2) atm II Final moles of gases in vessel I and II = Initial mole in vessel II or, P V R P V R \u00d7 \u00d7 + + \u00d7 \u00d7 600 2 600 ( ) = 4 300 \u00d7 \u00d7 V R \u21d2 P = 3 atm
Answer: 4
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Solution: Let the mixture contains a moles C x H 8 and b moles C x H 10 . Now, a \u00d7 (12 x + 8) + 6 \u00d7 (12 x + 10) = 28.4 (1) a + b = PV RT = 2 46 5 0 082 300 . . \u00d7 \u00d7 = 0.5 (2) and 28.4 \u00d7 84 5 100 . = a \u00d7 12 x + b \u00d7 12 x (3) On solving, x \u2248 4
Answer: 6
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Solution: From PV = nRT , V = nR P T . As the pressure is constant, the change in slope is only due to change in moles. X n \u2192 nX Initial a mole o Final a \u2013 0.6 a 0.6 axn = 0.4 a Now, a a an 0 4 0 6 . . + = ( . . ) / ( . . ) / 50 2 49 9 20 49 1 47 9 20 \u2212 \u2212 \u21d2 n = 6
Answer: 4
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Solution: 2H 2 + O 2 \u2192 2H 2 O 2a a 0 Final 2a \u2013 1.6a a \u2013 0.8a 1.6a = 0.4a = 0.2a Now, P nT = R V = Constant \u21d2 P n T 1 1 1 = P n T 2 2 2 \u21d2 4 5 3 330 . a \u00d7 = P a 2 2 2 400 . \u00d7 \u2234 P 2 = 4 atm
Answer: 8
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Solution: n total = n n O N 2 2 + or, 1 1 30 . \u00d7 RT = Po RT RT 2 30 0 9 10 \u00d7 + \u00d7 . \u21d2 P O 2 = 0.8 atm
Answer: 4
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Solution: P H O 2 = 40 100 \u00d7 (V.P.)\n3.58 Chapter 3 HINTS AND EXPLANATIONS First drop of liquid will form when P = V.P. Now, P 1 V 1 = P 2 V 2 \u21d2 40 100 \u00d7 (V.P.) \u00d7 10 = (V.P.) \u00d7 V 2 \u2234 V 2 = 4 ml
Answer: 9
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Solution: H 2 O(l) \u2192 H 2 (g) + 1 2 O 2 (g) a mole 0 0 Final 0 a mole 0.5 a mole \u0394 P.V = \u0394 n .RT or, (1.86 \u2013 0.96) \u00d7 20 = (1.5a) \u00d7 0.08 \u00d7 300 \u21d2 a = 0.5 \u2234 Mass of water present initially = 0.5 \u00d7 18 = 9 gm
Answer: 4
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Solution: Initial total moles = PV RT = 24 63 3 0 0821 300 . . \u00d7 \u00d7 = 3 \u2234 Initial mole of H 2 = 3 \u2013 1 = 2 Final mole ratio, n n H D 2 2 1 2 4 4 1 2 = = / / Now, n n n n M M f f i i n H D H D D H 2 2 2 2 2 2 = \u00d7 \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f or, 1 2 2 1 4 2 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f n \u21d2 n = 4
Answer: 2
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Solution: For critical point, dP dV m = 0 and d P dV m 2 2 0 = On solving, b =
Answer: 8
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Solution: But b \u2260 0 from question. Hence, the gas does not have critical condition.
Answer: 0775
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Solution: Boyle\u2019s temperature: T a Rb B = = \u00d7 = 4 105 0 0821 0 1 500 . . . K Hence, at 500 K, the gas will behave ideally. Not, d PM RT = = \u00d7 \u00d7 2 164 2 0 0821 500 . . = 8 g/L = 8 kg/m 3 Four-digit Integer Type
Answer: 3900
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Solution: l mm 760 750 l \u2013 750 800 770 l \u2013 770 P 760 l \u2013 760 (760 \u2013 750) \u00d7 ( l \u2013 750) = (800 \u2013 770) \u00d7 ( l \u2013 770) = ( P \u2013 760) \u00d7 ( l \u2013 760) \u2234 P = 775 mm Hg
Answer: 0404
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Solution: Mass of LNG = 10 m 3 \u00d7 416 kg/m 3 = 416 \u00d7 10 4 gm \u2234 Moles of CH 4 = 916 10 16 4 \u00d7 = 26 \u00d7 10 4 Now, V = nRT P = \u00d7 \u00d7 \u00d7 26 10 0 021 300 1 692 4 . . = 3.9 \u00d7 10 6 L = 3900 m 3
Answer: 8724
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Solution: V n T V n T V n V n T 1 1 1 2 2 2 2 303 1 6 1 2 = \u21d2 \u00d7 = \u00d7 . . \u21d2 T 2 = 404 K
Answer: 0450
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Solution: V initial = a = nRT P = \u00d7 \u00d7 \u00d7 64 0 08 300 64 3 . = 8 L V initial = b = nRT P = \u2212 \u00d7 \u00d7 \u00d7 ( ) . 64 8 0 08 300 64 3 = 7 L P = nRT V = \u00d7 \u00d7 \u00d7 = 64 0 08 300 64 7 24 7 . atm\n3.59 Gaseous State HINTS AND EXPLANATIONS
Answer: 2421
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Solution: P 1 320 K P 0 P 2 T K P 0 P \u2032 2 P \u2032 1 P 2 + P 0 = P 1 P \u2019 2 + P 0 = P \u2019 1 P 0 = P 1 \u2013 P 2 = P \u2019 1 \u2013 P \u2019 2 or, n R V n R V n R T V n R T V \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 \u00d7 \u239b \u239d \u239c 320 5 320 4 5 4 3 4 \u239e \u239e \u23a0 \u239f \u2234 T = 450 K
Answer: 1254
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Solution: Mass of gas used = 28.8 \u2013 23.2 = 5.6 kg Volume of gas used up, V = nRT P = \u00d7 \u00d7 \u00d7 \u00d7 ( . ) . 5 6 10 0 08 300 56 1 3 = 2400 L Now, P M P M P 1 1 2 2 2 35 28 8 14 8 23 2 14 8 = \u21d2 \u2212 = \u2212 ( . . ) ( . . ) \u21d2 P 2 = 21 atm
Answer: 0065
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Solution: n NO = \u00d7 \u00d7 1 6 0 75 0 08 3 00 . . . . = 0.05 n O 2 1 2 0 25 0 08 3 00 = \u00d7 \u00d7 . . . . = 0.0125 2NO + O 2 \u2192 2NO 2 \u2192 N 2 O 4 0.05 0.0125 0 0 Final \u2212 0 025 0 025 . . \u2212 0 0125 0 . 0 0 0 0125 0 0125 . . But at 200 K, N 2 O 4 is solid. Hence, the only gas is NO. Millimoles of NO remained = 0.025 \u00d7 1000 = 25 Now, P = 0 025 0 08 200 0 75 0 25 . . ( . . ) \u00d7 \u00d7 + = 0.4 atm
Answer: 0500
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Solution: 76 70 6 76 l = ? 3 P 1 = (6 \u2013 1) = 5 cm Hg P 2 = ? V 1 = 6 A cm 3 V 2 = 3 A cm 3 P 2 = 5 6 3 \u00d7 A A = 10 cm Hg Hence, fi nal total pressure in table above mercury = 10 + 1 = 11 cm Hg \u2234 Barometer reading = 76 \u2013 11 = 65 cm
Answer: 0320
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Solution: Mass of water vapour present initially, m V R 1 756 100 24 760 300 18 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 Mass of water vapour fi nally remained, m V R 2 8 4 760 280 18 = \u00d7 \u00d7 \u00d7 . \u2234 Fraction of water condensed = m m m 1 2 1 \u2212 = 0.5
Answer: 3610
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Solution: Let the process time = t min Mass of water vapour in inlet air, m t 1 3 20 100 38 760 10 10 0 08 500 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 ( ) \u00d7 \u00d7 . 18 gm = 2.5 t gm Mass of water vapour in outlet air, m t 2 3 80 100 19 760 10 10 0 08 400 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 ( ) \u00d7 \u00d7 . 18 gm = 6.25 t gm From question, m 1 + 200 kg \u00d7 36 100 = m 2 \u21d2 t = 19200\n3.60 Chapter 3 HINTS AND EXPLANATIONS
Answer: 1836
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Solution: At a depth of 10 m, P 1 = 1 atm + 10 m water = 1 + 1 013 1000 1000 1 013 10 6 . . \u00d7 \u00d7 \u00d7 = 2 atm V 1 = 24 ml n 1 = 2 24 10 0 08 300 3 \u00d7 \u00d7 \u00d7 \u2212 . = 2 \u00d7 10 \u2212 3 At surface, P 2 = 1 atm, V 2 = ?, n 2 = 2 \u00d7 10 \u2212 3 \u2212 0.05 \u00d7 10 \u2212 3 \u00d7 10 = 1.5 \u00d7 10 \u2212 3 Now, PV n P V n V V 1 1 1 2 2 2 3 2 3 2 2 24 2 10 1 1 5 10 = \u21d2 \u00d7 \u00d7 = \u00d7 \u00d7 \u21d2 \u2212 \u2212 . = 36 ml For volume remaining uncharged, P n P n 1 1 2 2 = or, 2 2 10 1 2 10 10 10 3 3 4 \u00d7 = \u00d7 \u2212 \u00d7 \u21d2 = \u2212 \u2212 \u2212 r r mol/min
Answer: 0013
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Solution: N 2 O 4 \u2192 2NO 2 Initial mole 20 \u00d7 V RT 0 Final mole 20 10 \u00d7 \u2212 V RT V RT 20 V RT = 10 V RT NO 2 will eff use through SPM till its pressure becomes same in both chamber and hence, mole ratio of NO 2 in chamber-I and II should be 1 :
Answer: 0140
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Solution: Final moles in chamber-I = 10 V RT of N 2 O 4 and 5 V RT of NO 2 Final moles in chamber-II except H 2 O vapour = 15 V RT of NO 2 \u2234 Pressure of gas in chamber-I = 15 1 2 V RT R T V \u00d7 \u00d7 . = 18 mm and pressure of gases in chamber-II = 15 1 2 3 V RT R T V \u00d7 \u00d7 . + 30 = 36 mm
Answer: 0015
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Solution: 5 2 n \u2264 0.01 \u21d2 n \u2265 12.28
Answer: 0050
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Solution: H2 H2 H2 H2 O2 N2 N2 N2 Inital 30 mole 5 mole 5 mole O2 = 5 mole N2 = 2.5 mole N2 = 2.5 mole Final H2 = 10 mole H2 = 10 mole H2 = 10 mole P 1 : P 2 : P 3 = 10 : 17.5 : 12.5 = 4 : 7 : 5
Answer: 9450
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Solution: A1 = n 2 A1 = n 2 A3 = n 4 A3 = n 4 A3 = n 4 A2 = n 3 A2 = n 3 A3 = n 4 A2 = n 3 \u2234 P P n n N A A N 4 1 5 \u2212 = / / = 3 \u21d2 N = 15
Answer: 0100
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Solution: r r M M t t M H air air H air 2 2 100 26 2 = \u21d2 = / / and r r M M t t M M gas air air H air gas 2 = \u21d2 = 100 130 / / \u2234 M gas = 50
Answer: 0250
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Solution: 3 2 kT = mgh \u21d2 3 2 \u00d7 8 4 . N A \u00d7 300 = ( ) 40 10 3 \u00d7 \u2212 N A \u00d7 10 \u00d7 h \u2234 h = 9450 m
Answer: 0096
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Solution: At Boyle\u2019s temperature, the second virial coefficient is zero. B = a + b . e c T / 2 = 0 or, e c T \u2212 / 2 = \u2212 a b \u21d2 e T \u2212 950 2 / = \u2212 \u2212 0 02 0 22 . . \u21d2 T = 100 k\n3.61 Gaseous State HINTS AND EXPLANATIONS
![\"Ionic](\"./assets/chapter-packs/ionic-equilibrium/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: [ ] NH M 2 30 15 10 10 \u2212 \u2212 \u2212 = = \u2234 Number of NH 2 \u2212 ions per ml = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 10 1 1000 6 10 6 10 15 23 5 ( )
Answer: D
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Solution: On increasing temperature, the dissociation of water will increase. It will result increase in [H + ] and as well as in [OH \u2013 ] and hence, decreases in P H and as well as P OH .
Answer: B
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Solution: For maximum dissociation, [H + ] = [OH \u2013 ].
Answer: B
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Solution: [H + ] = 10 \u20132 M \u21d2 n H + = \u00d7 = \u00d7 \u2212 \u2212 200 10 1000 2 10 2 3 [OH \u2013 ] = 10 \u20132 M \u21d2 n OH \u2212 = \u00d7 = \u00d7 \u2212 \u2212 300 10 1000 3 10 2 3 \u2234 Moles of excess OH \u2013 remained = 1 \u00d7 10 \u20133 [OH \u2013 ] = 1 10 500 1000 2 10 3 3 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 M \u2234 P OH = \u2013 log (2 \u00d7 10 \u20133 ) = 2.7 \u21d2 P H = 11.3
Answer: A
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Solution: [OD \u2013 ] excess = 80 0 1 20 0 2 100 0 04 \u00d7 \u2212 \u00d7 = . . . M \u2234 P OD = \u2013 log (0.04) = 1.4 Now, P Kw of D 2 O = P D + P OD = 13.6 + 1.4 = 15 \u2234 Kw = 1 \u00d7 10 \u201315
Answer: D
\nOriginal PDF solution page
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Solution: [OT - ] excess = 400 0 2 100 0 4 500 0 08 \u00d7 \u2212 \u00d7 = . . . M \u2234 P OT = \u2013 log(0.08) = 1.1 Now, PT = P Kw \u2013 P OT = 2 \u00d7 7.60 \u2013 1.1 = 14.1
Answer: A
\nOriginal PDF solution page
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Solution: K Kw Ka b = = \u00d7 = \u00d7 \u2212 \u2212 \u2212 10 2 10 5 10 14 10 5
Answer: B
\nOriginal PDF solution page
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Solution: NH + H O NH OH 3 2 \u001f \u21c0 \u001f \u21bd \u001f \u001f 4 + \u2212 + \u0394 H \u00b0 = (\u201352.21) + (54.70) = 2.49 kJ \u0394 S \u00b0 = 1.6 + (\u201376.3) = \u2013 74.7 J/K Now, \u0394 H \u00b0 = \u2013 RT. ln K eq or, 2490 \u2013 300 \u00d7 (\u201374.7) = \u20138.3 \u00d7 300 \u00d7 ln K eq \u2234 K eq = e \u201310
Answer: A
\nOriginal PDF solution page
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Solution: [ ] ] H [H HCOOH CH COOH 3 + + = or, 2 4 10 0 6 1 8 10 8 4 5 . . . \u00d7 \u00d7 = \u00d7 \u00d7 \u21d2 = \u2212 \u2212 C C M \u2234 Moles of CH 3 COOH added = 100 8 1000 0 8 \u00d7 = .
Answer: D
\nOriginal PDF solution page
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Solution: K a ( HA ) = K b ( A \u2013 ) = Kw = \u2212 10 7 Now, [ ] . . H M P H + \u2212 \u2212 = \u00d7 = \u21d2 = 10 0 1 10 4 0 7 4
Answer: A
\nOriginal PDF solution page
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Solution: NH +OH K M S K NH +H O 4 + f b 3 2 \u2212 \u2212 \u2212 = \u00d7 = 3 4 10 10 1 1 . ? Given : NH NH H K M 4 + 3 \u001f \u21c0 \u001f \u21bd \u001f \u001f + = \u00d7 + \u2212 ; . 1 10 5 6 10 and H O H OH K M 2 \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 \u2212 + = \u00d7 ; . 2 14 2 1 0 10 \u2234 NH OH NH H O K K K 4 + 3 2 eq + + = \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f ; 1 2 Now, 3 4 10 5 6 10 10 6 07 10 10 10 14 5 1 . . . \u00d7 = \u00d7 \u21d2 = \u00d7 \u2212 \u2212 \u2212 K K S b b
Answer: B
\nOriginal PDF solution page
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Solution: CH COOH CH COO H 3 3 0 1 0 1 0 1 . . . \u2212 + + \u2248 + \u2212 + + x x y y x \u001f \u21c0 \u001f \u21bd \u001f \u001f Cl CHCOOH Cl CHCOO H 2 2 0 1 0 1 0 1 . . . \u2212 + + \u2248 + \u2212 + + y x y y y \u001f \u21c0 \u001f \u21bd \u001f \u001f 0 15 0 1 0 1 0 05 . ( . ) ( . ) . = \u00d7 + \u2212 \u21d2 = y y y y \u2234 [H + ] = 0.1 + x + y \u2248 0.1 + y = 0.15 M \u2234 P H = \u2013 log(0.15) = 0.82
Answer: A
\nOriginal PDF solution page
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Solution: [ ] . . / / . OH M \u2212 = \u00d7 = 0 4 100 4 25 17 250 1000 0 004 P OH = \u2013 log(0.004) = 2.4 \u2234 P H = 14 \u2013 2.4 = 11.6
Answer: B
\nOriginal PDF solution page
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Solution: [ ] . . OH K C M b \u2212 \u2212 \u2212 = \u00d7 = \u00d7 \u00d7 = \u00d7 1 6 10 0 0025 4 10 6 9 P P OH H = \u2212 \u00d7 = \u21d2 = \u2212 log . . 4 10 4 2 9 8 9 EXERCISE II (JEE ADVANCED)\n7.42 Chapter 7 HINTS AND EXPLANATIONS
Answer: A
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Solution: [ ] / O HSaC M = \u00d7 = \u00d7 \u2212 \u2212 4 10 200 1000 2 10 4 3 and P H = 3.0 \u21d2 [H + ] = 10 \u20133 M Now, 2 10 10 2 10 4 10 12 3 3 12 \u00d7 = \u00d7 \u00d7 \u21d2 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 [S ] [ ] aC SaC M
Answer: B
\nOriginal PDF solution page
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Solution: [ ] . . HA M O = \u00d7 = 20 0 5 50 0 2 [ ] . . HB M O = \u00d7 = 30 0 2 50 0 12 HA H A 0 2 . \u2212 + + \u2212 + x x y x \u001f \u21c0 \u001f \u21bd \u001f \u001f HB H 0 12 . \u2212 + + \u2212 + y x y y B \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 2 10 0 2 0 2 4 \u00d7 = + \u22c5 \u2212 \u2248 + \u22c5 \u2212 ( ) x . ( ) . x y x x y x \u2234 ( x + y ) \u22c5 x = 4 \u00d7 10 \u20135 (1) and, 5 10 0 12 0 12 5 \u00d7 = + \u22c5 \u2212 \u2248 + \u22c5 \u2212 ( ) ( . ) ( ) . x y y y x y y \u2234 ( x + y ) \u22c5 y = 6 \u00d7 10 \u20136 (2) From (1) and (2), [H + ] = x + y = 6.78 \u00d7 10 \u20133 M
Answer: C
\nOriginal PDF solution page
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Solution: 10 0 01 10 5 8 \u2212 \u2212 = \u00d7 \u21d2 = K K a a . Now, [ ] . . . OH P OH \u2212 \u2212 \u2212 = \u00d7 = \u21d2 = 10 0 1 10 4 5 8 4 5 \u2234 P H = 9.5
Answer: B
\nOriginal PDF solution page
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Solution: RNH +H O RNH + OH 2 2 0 01 3 10 4 . \u2212 + \u2212 + \u2212 x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 10 10 0 01 10 6 4 4 \u00d7 = + \u2212 \u21d2 = \u2212 \u2212 \u2212 x x x x ( ) . \u2234 [OH \u2013 ] = 2 \u00d7 10 \u20134 M
Answer: C
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Solution: As on adding HCl, [H + ] is not changing and will remain unchanged.
Answer: A
\nOriginal PDF solution page
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Solution: Co +H O HCo H 2 2 Decrease Shiftleft \u2193 \u2190 \u2212 + + \u001f \u21c0 \u001f\u001f\u001f\u001f \u21bd \u001f \u001f\u001f\u001f\u001f 3 As [H + ] decreases, P H increases.
Answer: C
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Solution: [ ] . / / . W H M O 2 4 0 16 32 500 1000 0 01 = = \u2234 \u221d = 4 10 0 01 0 02 2 6 \u00d7 = \u2212 . . % or
Answer: B
\nOriginal PDF solution page
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Solution: P H = \u2013 log (2 \u00d7 10 \u20136 ) = 5.70
Answer: B
\nOriginal PDF solution page
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Solution: [OH\u2013] = 6.67 \u00d7 10 \u20133 + 6 67 10 2 0 10 3 2 . \u00d7 + \u2248 \u2212 \u2212 M \u2234 P OH = \u2013 log (10 \u20132 ) = 2.0 \u21d2 P H = 12.0
Answer: D
\nOriginal PDF solution page
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Solution: en + H O enH OH M M M b 2 0 09 5 1 8 1 10 ( . ) ( ) ( ) ;K . \u2212 + \u2212 \u2212 + \u2212 + = \u00d7 x x y x y \u001f \u21c0 \u001f \u21bd \u001f \u001f enH + H O enH OH M M M b + \u2212 + \u2212 + \u2212 + = \u00d7 2 2 2 8 2 7 0 10 ( ) ( ) ;K . x y y x y \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 8 1 10 0 09 0 09 2 7 10 5 3 . ( )( ) ( . ) . . \u00d7 = \u2212 + \u2212 \u2248 \u22c5 \u21d2 = \u00d7 \u2212 \u2212 x y x y x K x x x and 7 0 10 7 0 10 8 8 . ( ) ( ) . \u00d7 = \u22c5 + \u2212 \u2248 \u22c5 \u21d2 = \u00d7 \u2212 \u2212 y x y x y y x x y \u2234 [en H + ] = ( x \u2013 y ) \u2248 x M = 2.7 \u00d7 10 \u20133 M [enH ]= = 7.0 10 M 2 2+ y \u00d7 \u2212 8 \u2234 [OH \u2013 ] = ( x + y ) = x = 2.7 \u00d7 10 \u20133 M and P OH = \u2013 log (2.7 \u00d7 10 \u20133 ) = 2.56 \u21d2 P H = 11.44
Answer: B
\nOriginal PDF solution page
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Solution: K K H H S a a 2 1 2 2 2 5 \u22c5 = + \u2212 [ ] [ ] [ ] or, ( . ) ( . ) ( . ) [ ] . [ ] . 1 4 10 1 0 10 0 1 5 0 2 5 2 8 10 7 14 2 2 2 20 \u00d7 \u00d7 \u00d7 = \u00d7 \u21d2 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 M
Answer: A
\nOriginal PDF solution page
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Solution: [ ] . H M + \u2212 \u2212 = \u00d7 \u00d7 = \u00d7 0 2 2 10 2 10 5 3 Now, ( ) ( ) ( ) ( ) [ ] [ ] 2 10 5 10 4 10 2 10 5 9 12 3 3 3 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 + A H A 3 \u2234 = \u00d7 \u2212 \u2212 [ ] [ ] A H A 3 3 17 5 10
Answer: D
\nOriginal PDF solution page
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Solution: [ ] . H M + \u2212 \u2212 = \u00d7 = 0 1 10 10 5 3 Now, K a 3 3 2 3 2 13 3 10 10 10 10 = \u21d2 = = + \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 [ ][ ] [ ] [ ] [ ] H A HA A HA \u2234 P X = 10
Answer: C
\nOriginal PDF solution page
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Solution: OH ACOH ACO H O m mol Final 0 m mol 1 m mol m mol \u2212 \u2212 + + 2 3 0 2 2 \u001f \u21c0 \u001f \u21bd \u001f \u001f p H = + = 4 74 2 1 5 04 . l og . (Acidic) Addition of 1 ml ACOH will decrease P H by 0.3 unit.\n7.43 Ionic Equilibrium HINTS AND EXPLANATIONS
Answer: A
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Solution: Millimoles of ACOH = 6 \u00d7 0.1 = 0.6 Millimoles of ACO \u2013 =
Answer: A
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Solution: \u00d7 0.1 = 1.2 \u2234 = + = p H 4 75 1 2 0 6 5 05 . log . . .
Answer: B
\nOriginal PDF solution page
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Solution: 1st solution will finally have 1 mole of CH 3 COOH. \u2234 = P P H Ka 1 1 2 and for 2nd solution, P P H Ka 2 =
Answer: D
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Solution: P P H H 2 1 0 6 = + . p M C P /M C K K a a + = + + log / log log . y x 3 98 \u2234 = y x 3 98 .
Answer: C
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Solution: 4 0 5 0 0 5 0 05 1 1 . . log . . = + \u21d2 = C C M 6 0 5 0 0 5 5 0 2 2 . . log . . = + \u21d2 = C C M Now, fi nal P 0.05 + 5.0 0.5 + 0.5 H = + \u00d7 \u00d7 \u00d7 \u00d7 = 5 0 5 7 . log . V V V V
Answer: A
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Solution: For maximum \u03b2 \u03b1\u03b2 \u03b1 , [ ] H P P H K a + = \u21d2 = 0
Answer: A
\nOriginal PDF solution page
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Solution: BOH H B C Final C-0.1 V 0.1 40 40 40 40 40 0 0 0 1 40 \u00d7 + + + \u00d7 + + + V M V V V M V \u001f \u21c0 \u001f \u21bd \u001f \u001f . + + + V H O 2 H + must be a limiting reagent because both P H are above > .0. Now, P =P +log 0.1V 40C 0.1V OH K b \u2212 14 =P +log 0.1 5 40C 0.1 5 K b \u2212 \u00d7 \u2212 \u00d7 10 (1) 14 =P +log 0.1 40C 0.1 K b \u2212 \u00d7 \u2212 \u00d7 9 20 20 (2) \u2234 K b = 2 \u00d7 10 \u20135
Answer: B
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Solution: For maximum buffer capacity: [ [ [ ACOH] NaOH] NaOH]= M M = \u21d2 = 2 1 2 2 1 \u2234 Mass of NaOH added = \u00d7 \u00d7 = 500 1 1000 40 20 gm
Answer: B
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Solution: n n HA OH M M= 80 = \u21d2 = \u00d7 \u21d2 \u2212 0 28 35 0 1 1000 . .
Answer: C
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Solution: Fe H O Fe(OH H M 2 M 3 0 9 2 0 1 + + + + + . . ? ) x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 9 10 0 1 0 9 0 081 1 08 3 \u00d7 = \u00d7 \u21d2 = \u21d2 = \u2212 + + . [ ] . [ ] . . x x H H P H
Answer: A
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Solution: In fi nal solution: [HA] = [A \u2013 ]
Answer: D
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Solution: [ ] NH K K K C = 8.33 10 M w a b 3 4 = \u22c5 \u00d7 \u00d7 \u2212
Answer: C
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Solution: For equivalence point, 2 5 2 5 2 15 . \u00d7 = \u00d7 V Hcl \u2234 V HCl = 7.5 ml BOH + H B +H M Eqn. M + M 0 XM + M (0.1 )M 2 5 2 5 10 0 7 5 2 15 10 0 0 1 . . . \u00d7 \u00d7 \u2212 x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 2 aq O; K = = \u2212 \u2212 10 10 10 12 14 2 100 0 1 2 7 10 2 = \u2212 \u22c5 \u21d2 = \u00d7 = \u2212 + . . ( ) x x x x M H
Answer: D
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Solution: At 2nd equation point: P P +P H K K 2 3 = = + = 1 2 8 12 12 10 ( ) a a Now, K K K A H A 3 a a a 1 2 3 3 \u22c5 \u22c5 = + \u2212 [H ][ ] [ ] or 7 5 10 10 10 10 4 8 12 10 3 3 . ( ) [ ] [ ] \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 A H A 3 \u2234 = = \u00d7 \u2212 \u2212 \u2212 [ ] [ ] . . H A A 3 3 6 7 10 7 5 1 33 10
Answer: B
\nOriginal PDF solution page
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Solution: CO H HCO K M 100% run 0 M 0 M 3 2 0 35 0 35 3 0 0 35 11 1 4 10 \u2212 + \u2212 \u2212 + = \u00d7 . . . ; \u001f \u21c0 \u001f \u21bd \u001f \u001f For HCO 3 \u2212 solution, [ ] , . H K K M + \u2212 = = \u00d7 a a 1 2 1 4 10 8 Now, K H O HCO a 2 3 2 3 = + \u2212 \u2212 [ ][C ] [ ] \u2234 = \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u2212 \u2212 [ ] . . CO M 3 2 11 8 3 4 10 0 35 1 4 10 10\n7.44 Chapter 7 HINTS AND EXPLANATIONS
Answer: D
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Solution: [ ] . . . LaC M \u2212 = \u00d7 = 0 0 125 0 5 2 0 5 Now, P P C OH K a = \u2212 + 7 1 2 ( log ) 5 6 7 1 2 0 5 . ( log . ) = \u2212 + P K a \u2234 = \u21d2 = \u00d7 \u2212 P K K a 3 1 8 10 4 . a
Answer: A
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Solution: As HA is stronger acid, it will react first. For first equivalent point, V V ml NaOH NaOH \u00d7 = \u00d7 \u21d2 = 0 2 50 0 05 12 5 . . . At fi rst equivalent point, [ ] . . . A M \u2212 = \u00d7 = 50 0 05 62 5 0 04 [ . . . HB]= M 50 0 08 62 5 0 064 \u00d7 = Now, A HB B HA ; K = K (HB) K (HA eq a a \u2212 \u2212 \u2212 \u2212 + + 0 04 0 04 0 064 0 064 . . . . x x x x \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f ) ) = = = \u00d7 \u2212 \u2212 \u2212 \u2212 10 10 10 4 10 8 2 3 8 4 4 5 . . . 4 10 0 04 0 064 3 2 10 5 4 \u00d7 = \u22c5 \u00d7 \u21d2 = \u00d7 \u2212 \u2212 x x x . . . Now, K (HA H A HA H a ) [ ][ ] [ ] . [ ] . . = \u21d2 \u00d7 = \u00d7 \u00d7 + \u2212 \u2212 + \u2212 1 6 10 0 04 3 2 10 4 4 \u2234 = \u00d7 \u21d2 = + \u2212 [ ] . . H P H 1 28 10 5 9 6
Answer: C
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Solution: CrO +H O HCrO +OH ; K Kw K 4 2 2 4 h \u2212 \u2212 \u2212 \u2212 \u2212 = = \u00d7 0 005 8 2 2 10 . x x x a \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 10 0 005 0 005 10 8 2 5 \u00d7 = \u22c5 \u2212 \u2248 \u21d2 = \u2212 \u2212 x x x x x . . \u2234 = = \u2212 h 10 0 005 0 002 5 . .
Answer: A
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Solution: P and P K K a a 1 2 2 40 9 60 = = . . \u2234 Required pH = + = 1 2 2 40 9 60 6 00 ( . . ) .
Answer: C
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Solution: HA H A Red Blue \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + [ ] [ ] [ ] H K HA A + \u2212 = \u22c5 a \u2234 = \u2212 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f + + + [ ] [ ] [ ] H required H H K a 2 1 75 25 25 75 = 8 \u00d7 10 \u20135 M
Answer: A
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Solution: 5 5 4 75 . . log [ ] [ = + \u2212 ACo ACOH] O O \u2234 = \u2212 [ ] [ . ACo ACOH] O O 5 62 1
Answer: C
\nOriginal PDF solution page
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Solution: n n CO H 2 and used = = = \u00d7 = + 224 22400 0 01 30 1 1000 0 03 . . CO +OH HCO 2 mole (L.R. 3 mole or less \u2212 \u2212 \u23af \u2192 \u23af 0 01 0 01 . ) . Hence, moles of H + used should be 0.01 or less and titration of HCO 3 \u2212 and H+ should not be detected by phenolphthalein. Hence, OH \u2013 must be in excess. CO + 2OH CO H O 2 mole mole 3 mole 2 0 01 0 02 2 0 01 . . . \u2212 \u2212 \u23af \u2192 \u23af + Thus, 0.01 mole of CO 3 2 \u2212 will require only 0.01 mole of H + in the presence of phenolphthalein. As the mole of H + used is 0.03, 0.02 mole OH \u2013 must be present in excess. Hence, total moles of OH \u2013 used = 0.02 + 0.02 = 0.04, \u2234 = = [ ] . . NaOH M used 0 04 1 0 04
Answer: A
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Solution: PbSO g/100ml 4 = \u00d7 \u00d7 \u00d7 \u2212 \u2212 2 10 304 10 1 36 10 9 3 \u001b . ZaS g/100ml = \u00d7 = \u00d7 \u2212 \u2212 10 97 10 9 7 10 22 11 . AgBr g/100ml = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 4 10 188 10 1 19 10 13 5 . CuCo g/100ml 3 8 3 10 123 10 1 23 10 = \u00d7 = \u00d7 \u2212 \u2212 .
Answer: C
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Solution: Hg 4Cl HgC 2+ M 100% ram 0 Eqn. 1.6 M M 0.5M 0.5M 0 1 10 0 9 17 . . \u00d7 \u2212 \u2212 + \u001f \u21c0 \u001f \u21bd \u001f \u001f l l M 0.1M 4 2 0 0 1 \u2212 . \u2234 = \u00d7 \u00d7 = \u2212 K form 0 1 1 6 10 0 5 10 17 4 17 . . ( . )\n7.45 Ionic Equilibrium HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: AgBr(s) S O Ag(S O Eqn/final 0 aM a 0 1 2 3 2 0 2 2 3 2 3 0 0 1 2 . . . ) + + \u2212 \u2212 \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f B Br \u2212 0 0 1 . K eq = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u21d2 = \u2212 4 10 1 6 10 0 1 0 1 0 2 0 325 13 12 2 . . . ( . ) . a a
Answer: A
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Solution: Tl S Tl S SM 2 2 2 2 (S) \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + x S H O HS OH Kw K M SM SM h a 2 2 2 \u2212 \u2212 \u2212 + + = x \u001f \u21c0 \u001f \u21bd \u001f \u001f ; K 10 10 2 10 2 10 4 10 14 14 6 6 12 \u2212 \u2212 \u2212 \u2212 \u2212 = \u00d7 \u00d7 \u00d7 \u21d2 = \u00d7 x x \u2234 = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 K sp ( ) . 2 2 10 4 10 6 4 10 6 2 12 23
Answer: A
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Solution: M SCN M Eqn. M M 3 2 10 2 10 5 10 1 51 10 1 51 10 1 3 3 4 3 3 + \u00d7 \u00d7 \u2212 = \u00d7 \u2212 \u00d7 \u00d7 \u2212 = \u2212 \u2212 \u2212 \u2212 \u2212 + x x . . . . . ) 0 10 2 0 1 5 10 5 3 \u00d7 + = \u00d7 \u2212 \u2212 M M M M(SCN \u001f \u21c0 \u001f \u21bd \u001f \u001f x \u2234 = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 K f 1 5 10 5 10 1 10 3 10 3 4 5 5 .
Answer: B
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Solution: SrCO s) Sr CO M 3 2 2 10 3 2 2 10 4 4 ( \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u00d7 \u2212 \u00d7 \u2212 \u2212 \u2212 + x CO H O HCO OH M M M 3 2 2 10 2 3 4 10 4 6 \u2212 \u00d7 \u2212 \u2212 \u2212 \u00d7 \u2212 \u2212 + + ( ) x x \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 10 5 10 4 10 2 10 0 01 51 14 11 6 4 \u2212 \u2212 \u2212 \u2212 \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 \u21d2 = x x x ( ) . \u2234 = \u00d7 \u00d7 \u00d7 \u2212 = \u00d7 \u2212 \u2212 \u2212 K sp ( ) ( ) 2 10 2 10 4 51 10 4 4 8 x
Answer: C
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Solution: MnS(S) Mn S SM S M \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 2 + \u2212 \u2212 + ( ) x S H O HS OH S M M M 2 2 \u2212 \u2212 \u2212 \u2212 + + ( ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 10 10 14 14 10 \u2212 \u2212 \u2212 = \u22c5 \u2212 \u00d7 = \u22c5 \u2212 x x x x ( ) ( ) S and 2.5 10 S S \u2234 S = 6.3 \u00d7 10 \u20134 M
Answer: A
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Solution: AgCl(s) Br AgBr(s) Cl aq M M 0 1 0 075 0 075 0 075 . . . . (aq) ( \u2212 \u2212 \u2212 + + x \u001f \u21c0 \u001f \u21bd \u001f \u001f ) ) K Br K (AgCl) K (AgBr) Br eq sp sp = = \u21d2 = \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 [Cl ] [ ] . [ ] 0 075 2 10 4 10 10 13 3 \u2234 = \u00d7 \u2212 \u2212 [ ] . Br M 1 5 10 4
Answer: C
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Solution: [ ) ] . Ag(CN M 2 0 01 \u2212 = K Ag CN Ag CN Ag diss = \u21d2 \u00d7 = \u00d7 \u00d7 + \u2212 + \u2212 \u2212 + \u2212 [ ][ ] [ ( ) ] [ ] ( . ) . 2 2 20 7 2 1 10 2 5 10 0 01 \u2234 = \u00d7 + \u2212 [ ] . Ag M 1 6 10 9
Answer: B
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Solution: [ ] . CO M 3 2 2 0 \u2212 = Now, K (CaCO K (CaF CO F F F sp sp 3 2 3 2 2 3 3 4 2 8 ) ) [ ] [ ] [ ] ( ) = \u21d2 = \u21d2 = \u2212 \u2212 \u2212 \u2212 x y y x
Answer: C
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Solution: BaF s O BaC O s M M 2 2 4 2 0 1 2 4 2 2 ( ) C (aq) ( ) F (aq) ( . ) + \u21d2 + \u2212 \u2212 \u2212 x x K F C O K BrF K O eq 2 sp = = = = \u2212 \u2212 \u2212 \u2212 [ ] [ ] ( ) (BrC ) 2 4 2 2 4 6 10 4 10 10 10 sp \u2234 x \u2248 0.1 \u21d2 [F \u2013 ] = 0.2 M \u2234 [ ] ( . ) . Ba M 2 6 2 5 10 0 2 2 5 10 + \u2212 \u2212 = = \u00d7
Answer: C
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Solution: S Zn(OH) Zn(OH) Zn Zn(OH) Zn(OH) = + + + + + + \u2212 \u2212 [ (aq)] [ ] [ ] [ ] [ ] 2 2 3 4 2 = + \u22c5 + \u22c5 + + \u2212 \u2212 \u2212 \u2212 K K K OH K K K OH K K OH K K K [OH 5 4 1 1 2 1 3 2 1 2 4 1 2 [ ] [ ] [ ] ] = + \u00d7 + \u00d7 \u00d7 + \u00d7 \u00d7 + \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 10 10 10 0 1 10 10 10 0 1 10 10 0 1 10 10 6 7 6 4 7 6 2 3 6 . ( . ) . 3 3 6 2 10 0 1 \u00d7 \u00d7 \u2212 ( . ) = 10 \u20136 +10 \u201312 + 10 \u201315 + 10 \u20134 + 10 \u20134 \u2248 2 \u00d7 10 \u20134 M
Answer: D
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Solution: For molecular solubility of CaCl 2 , \u0394 H solution = 209.2 + (\u201333.5) = 175.7 KJ > 30 KJ For ionic solubility of CaCl 2 , \u0394 H solution = 209.2 + 1004.2 + 1715.4 \u2013 1598.3 \u2013 719.6 \u2013 711.2 = \u2013 100.3 KJ For molecular solubility of HgCl 2 , \u0394 H solution = 83.7 \u2013 66.9 = 16.8 KJ < 30 KJ For ionic solubility of HgCl 2 , \u0394 H solution = 83.7 + 460.2 + 2815.8 \u2013 1845.1 \u2013 719.6 \u2013 711.2 = 83.3 KJ > 30 KJ Hence, CaCl 2 is ionic and HgCl 2 is molecular solubility.
Answer: C
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Solution: [ ] . C O M 2 4 2 5 6 0 001 5 250 2 6 10 \u2212 \u2212 = \u00d7 \u00d7 \u00d7 = \u00d7 \u2234 = \u00d7 = \u00d7 \u2212 \u2212 K sp ( ) . 6 10 3 6 10 5 2 9\n7.46 Chapter 7 HINTS AND EXPLANATIONS
Answer: A
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Solution: Sr NO M Eqn. 0.001 M 0.05 M 2 0 001 0 001 75 100 3 0 05 0 05 + \u2212 = \u00d7 \u2212 \u2212 + . . . . x x \u001b \u001f \u21c0 \u001f \u001f \u21bd \u001f \u001f Sr(NO 3 0 ) + x \u2234 x = 0.00025 Now, K f = \u00d7 \u00d7 \u00d7 = \u2212 \u2212 2 5 10 7 5 10 0 05 20 3 4 4 . . .
Answer: A
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Solution: ACOAg(s) H Cl ACOH AgCl(s) Mole M M 0 1 0 1 0 1 . . . + + + + \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f K K K eq sp sp ACOH H Cl ACO ACO Ag Ag ACOAg] ( = \u00d7 \u00d7 = + \u2212 \u2212 \u2212 + + [ ] [ ][ ] [ ] [ ] [ ] [ ] [ A AgCl) \u00d7 K a = \u00d7 = \u21d2 \u2212 \u2212 \u2212 10 10 10 10 8 10 5 7 Almost complete reaction \u2234 = + \u2212 [ . , [H . ACOH] M ] =10 M \u001b 0 1 0 1 10 7 4 and [ ] [ ] [ ] . ACO Ka ACOH H M \u2212 + = \u00d7 = 0 01
Answer: B
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Solution: A B (s) A B x y y x x y \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + K x y s s K x y x y x y x y x y sp sp = \u22c5 \u22c5 \u21d2 = \u22c5 \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f + + 1 As K sp << 1, greater the value of ( x + y ), greater is s.
Answer: A
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Answer: C
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Solution: s = + = + + \u2212 \u2212 \u2212 [ ] [ [ ] [ ] Zn Zn(OH) K OH K OH sp f 2 4 2 2 For maximum or minimum S, d d OH S [ ] \u2212 = 0 or, \u2212 + = \u21d2 = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u2212 \u2212 2 2 0 10 3 1 4 4 K OH K OH OH M sp f sp [ ] [ ] [ ] K K f \u2234 P H = 10 and S min = 2.4 \u00d7 10 \u20139 M
Answer: A
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Solution: Al(OH) s) OH Al(OH) From question M 3 4 33 10 3 8 10 1 ( ; . ? + = \u00d7 \u2212 \u2212 \u2212 \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f K 6 6 10 50 34 \u00d7 = \u2212 50 10 2 10 9 30 3 5 = \u21d2 = \u00d7 \u21d2 = \u2212 \u2212 \u2212 \u2212 [OH ] [ ] . OH M P H As the calculated [OH \u2013 ] is minimum OH \u2013 , P H is minimum. Al(OH) (s) Al OH K 3 M \u001f \u21c0 \u001f \u21bd \u001f \u001f 3 33 10 3 3 8 10 + \u2212 \u2212 \u2212 + = \u00d7 ; 8 10 10 2 10 4 30 33 3 3 10 \u00d7 = \u00d7 \u21d2 = \u00d7 \u21d2 = \u2212 \u2212 \u2212 \u2212 \u2212 [ ] [ ] . OH OH M P H As the calculated [OH \u2013 ] is maximum OH \u2013 , P H is maximum.
Answer: C
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Solution: From the question, [ . Cu(CN) M 4 3 0 1 \u2212 = and [CN \u2013 ] = 0.2 M \u2234 = \u22c5 = \u00d7 \u00d7 = \u00d7 + \u2212 \u2212 \u2212 \u2212 [ ] [ ] [ ] . . ( . ) Cu Cu(CN) CN Instab K 4 3 4 15 4 6 4 10 0 1 0 2 4 10 1 13 M Now, [ ] [ ] . ( ) . S (Cu S) Cu M sp 2 2 2 27 13 2 2 2 56 10 4 10 1 6 10 \u2212 + \u2212 \u2212 \u2212 = = \u00d7 \u00d7 = \u00d7 K \u2234 = \u00d7 = \u00d7 \u00d7 \u00d7 = + \u2212 \u2212 \u2212 \u2212 [ ] [ ] [ ] . . . H H S S M 2 K a 2 21 2 10 1 6 10 0 1 1 6 10 10 and P H = 10.0
Answer: B
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Solution: S M ppm = \u00d7 = \u00d7 = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 1 6 10 4 10 4 10 136 10 10 4 136 5 3 3 3 6 . For increase in concentration 4 times, volume should be 1 4 th . Hence, 75 % water should be evaporated.
Answer: C
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Solution: During precipitation, the concentration of both Ba 2+ and SO 4 2 \u2212 ions will decrease.
Answer: B
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Solution: K sp AgCl) ( = \u00d7 = \u2212 \u2212 \u2212 10 10 10 4 6 10 K sp (Ag CrO 2 4 4 2 4 12 10 8 10 8 10 ) ( ) = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 After precipitation of AgCl, fi nd the concentration of Cl \u2013 . [ ] . Cl M final \u2212 \u2212 \u2212 \u2212 = \u00d7 \u2212 \u00d7 \u00d7 \u00d7 1 0 10 8 10 2 10 6 7 7 Now, [ ] [ ] ( CrO ) ( Cl) CrO Cl K Ag K Ag final final sp sp 4 2 2 4 2 \u2212 \u2212 = [ ] ( ) ( ) [CrO ] . CrO final final 4 2 7 2 12 10 2 4 2 2 10 8 10 10 3 2 1 \u2212 \u2212 \u2212 \u2212 \u2212 \u00d7 = \u00d7 \u21d2 = \u00d7 0 0 5 \u2212 M\n7.47 Ionic Equilibrium HINTS AND EXPLANATIONS Hence, moles of Ag 2 CrO 4 precipitated = \u00d7 \u2212 \u00d7 = \u00d7 \u2212 \u2212 \u2212 8 10 3 2 10 7 68 10 4 5 4 . .
Answer: A
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Solution: To prevent precipitation of AgCl, the concentration of Ag + needed in solution = K sp AgCl) Cl ( ( ) \u2212 = \u00d7 << \u2212 1 8 10 0 16 1 8 10 . . . M Hence, almost all Ag + ion must form complete with CN \u2013 ions. Ag CN Ag(CN) M CM M + \u00d7 \u2212 \u2212 \u2212 + = \u00d7 1 8 10 0 16 2 1 8 17 10 2 6 4 10 . . . ; . \u001f \u21c0 \u001f \u21bd \u001f \u001f K f
Answer: B
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Solution: [CO ] ( ) [ ] [ ] 3 2 2 3 2 \u2212 + = \u22c5 K overall H CO H a To prevent precipitation of MCO 3 , or, [ ][ ] M CO K sp 2 3 2 + \u2212 \u2264 or K K sp , [M ] [H CO ] [H ] 2 2 3 2 + + \u22c5 \u22c5 \u2264 a \u2234 \u2265 \u22c5 + + [H ] [ ] [ ] M K H CO K a sp 2 2 3 For MgCO 3 : [H ] . . . + \u2212 \u2212 \u2212 \u2265 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 0 1 5 10 0 05 9 10 2 5 10 17 8 6 M \u2234 P H \u2264 5.6 For SrCO 3 : [H ] . . + \u2212 \u2212 \u2212 \u2265 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 0 1 5 10 0 05 9 10 5 10 3 17 10 5 M \u2234 P H \u2264 4.78 For precipitation of SrCO 3 without any precipitation of MgCO 3 , the P H range should be 4.78 to 5.6
Answer: A
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Solution: Mn 2+ (aq) + H 2 S(aq) \u001c MnS(s) + 2H + (aq) To just start precipitation of MnS, Q < K eq or, [ ] [ ][ ] ) ( ) H Mn H S (H S MnS sp + + < 2 2 2 2 K K a or, [H ] . . . . [ ] + \u2212 \u2212 + \u2212 \u00d7 < \u00d7 \u00d7 \u21d2 < \u00d7 2 21 13 6 0 04 0 1 1 0 10 2 5 10 4 10 H M Now, in the given buffer, [ ] [ ] [ ] H CH COOH CH COO O O + \u2212 = \u22c5 K a 3 3 = \u00d7 \u00d7 = \u00d7 > \u00d7 \u2212 \u2212 \u2212 2 10 0 25 0 15 3 33 10 4 10 5 5 6 . . . M Hence, no precipitation. To start precipitation [H + ] should decrease and hence, CH 3 COONa should be added. Now, 4 10 2 10 0 25 6 5 \u00d7 = \u00d7 \u00d7 \u2212 \u2212 . [CH COONa] 3 O \u2234 [CH 3 COONa] O = 1.25 M
Answer: C
\nOriginal PDF solution page
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Solution: Mg(OH) S NH Mg NH OH 2 4 2 4 2 2 ( ) + + + + \u001f \u21c0 \u001f \u21bd \u001f \u001f To re-dissolve Mg(OH) 2 , Q \u2264 K eq or, [ ][ ] [NH ] Mg NH OH sp 2 4 2 4 2 2 + + \u2264 K K b or, 0 15 0 1 0 5 0 35 0 1 0 5 0 5 1 2 10 2 2 2 11 . . . . . . . . ( \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f \u2264 \u00d7 \u2212 n . . ) 0 10 5 2 \u00d7 \u2212 \u2234 n \u2265 0.035 Hence, minimum mass of (NH 4 ) 2 SO 4 needed = \u00d7 = 0 035 2 132 2 31 . . gm
Answer: B
\nOriginal PDF solution page
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Solution: Ag Cl M Eq 10 M + \u00d7 = \u00d7 \u00d7 \u2212 + \u2212 \u00d7 = \u00d7 \u2212 \u2212 + 500 0 01 1000 5 10 5 250 0 02 1000 5 3 3 . ( ) . x y 1 10 5 10 3 3 \u2212 \u2212 \u00d7 \u2212 M M AgCl(S) ( ) x \u001f \u21c0 \u001f \u21bd \u001f \u001f Ag Br AgBr( M 10 M M M + = \u00d7 \u00d7 \u2212 + \u2212 = \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 + 5 10 5 5 10 5 10 3 3 3 3 ( ) ( ) x y y \u001f \u21c0 \u001f \u21bd \u001f \u001f S S) As both reactions will tend towards completion, ( x + y ) = 5 \u00d7 10 \u20133 Now, [Ag ]( ) [ ] + \u2212 \u2212 + \u2212 \u00d7 \u2212 = \u21d2 \u22c5 = 5 10 10 10 3 10 10 x y Ag (1) and [Ag ]( ) + \u2212 \u2212 \u00d7 \u2212 = \u00d7 5 10 5 10 3 13 y (2) From (1) \u00f7 (2), y y y 5 10 200 1 201 3 \u00d7 \u2212 = \u21d2 = \u2212 \u2234 = \u00d7 \u2212 \u2248 \u00d7 \u2212 \u2212 \u2212 [ ] . Br M 5 10 2 5 10 3 5 y\n7.48 Chapter 7 HINTS AND EXPLANATIONS
Answer: D
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+ "title": "Chem Sec 2",
+ "originalName": "Section B - Multi Correct",
+ "questions": [
+ {
+ "question_id": "ionic-equilibrium-chem-sec-2-1-81",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": 1,
+ "subject": "chemistry",
+ "chapter": "ionic-equilibrium",
+ "chapterTitle": "Ionic Equilibrium",
+ "type": "mcqm",
+ "rawChapterType": "MSQ",
+ "originalNumber": 81,
+ "displayNumber": 1,
+ "image": "./assets/chapter-packs/ionic-equilibrium/Chemistry Section 1__--__81__--__1.png",
+ "solutionImage": null,
+ "question": {
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+ "identifier": "A",
+ "content": ""
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+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
+ "content": ""
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+ "identifier": "D",
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+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: (a) Complete neutralization \u21d2 P H = 7.0 (b) [ ] . . . . H M P final H + = \u00d7 \u2212 \u00d7 = \u21d2 = 55 0 1 45 0 1 100 0 01 2 0 (c) OH \u2013 is in excess. (d) [ ] . . H M P final H + = \u00d7 \u2212 \u00d7 = \u21d2 = 75 1 5 25 1 5 100 0 1 1 0
Answer: A, B, C
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Solution: For basic solution: [ ] [ ] ] H OH and [H Kw + \u2212 + < < \u2234 P H > P OH or P P or P P H kw OH kw > < 2 2
Answer: C
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Solution: HA H A C(1 M C M C M \u2212 + \u2212 + \u03b1 \u03b1 \u03b1 ) \u001f \u21c0 \u001f \u21bd \u001f \u001f K C C C C C K C a a = \u22c5 \u2212 = \u22c5 \u2212 \u2248 \u22c5 \u21d2 = \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 ( ) 1 1 2 2 Now, K C C K K a a a = \u22c5 \u2212 = \u22c5 \u2212 \u21d2 = + + + + [ ] ( ) [ ] [ ] H H H \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 1 1 = + = + + \u2212 1 1 1 1 10 [ ] ( ) H P P Ka H K a
Answer: A, B, C
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Solution: Dilution results in increased degree of dissociation but decrease in concentrations of all active components.
Answer: A, B
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Solution: Relation is valid only for conjugate pairs.
Answer: B
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Solution: P H may decrease only on increasing [H + ].
Answer: A, B, C, D
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Solution: CH COOH CH COO H M 0.1 M M M 0.1 M 3 3 0 1 0 1 ( . ) ( . ) \u2212 + \u2212 + + x x x \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 1 8 10 0 1 0 1 1 8 10 5 5 . . . . \u00d7 = \u00d7 \u21d2 = \u00d7 \u2212 \u2212 x x and \u03b1 = = \u00d7 \u2212 x 0 1 1 8 10 4 . . Now, [ ] [ ] [ ] H OH Kw H M T from water acid = = = \u2212 + \u2212 10 13
Answer: A, B, C, D
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Solution: RNH g) H O(l) RNH aq OH aq bar M M 2 1 2 3 ( ( ) ( ) + + + \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f x x 10 1 10 3 0 11 0 6 3 \u2212 \u2212 = \u22c5 \u21d2 = \u21d2 = \u21d2 = x x x P P OH H . .
Answer: D
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Solution: NH OH(aq) NH OH 4 4 \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + Addition of solid NH 4 OH will increase NH 4 OH(aq) concentration and hence, [OH \u2013 ] will increase.
Answer: B, C
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Solution: (c) H O H O H O OH ve 2 2 3 + + \u0394 \u00b0 = + + \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f ; H (d) HA OH A H O Final a a a a 2 2 0 2 2 + + \u2212 \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f P P P H K K a a = + = log / / a a 2 2
Answer: D
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Solution: [ ][ ] H C O + \u2212 >> 3 2
Answer: A, B, D
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Answer: A, C
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Solution: NH Cl NaOH NH OH NaCl For buffer: ( 4 4 0 0 a a b b a b b \u2212 \u2248 \u21d2 > + + ) \u001f \u21c0 \u001f \u21bd \u001f \u001f CH COONa HCl CH COOH NaCl For buffer: ( 3 3 0 0 a a b b a b b \u2212 \u2248 \u21d2 > + + ) \u001f \u21c0 \u001f \u21bd \u001f \u001f
Answer: A, B, C, D
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Solution: KCN is a salt of weak acid (HCN) and strong base (KOH).
Answer: B, C
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Solution: BOH H B mole th run a mole Equivalent point a a b a 1 5 5 0 0 0 0 \u2212 + + + \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f 5 5 2 a mole H O + Now, for 1/5th reaction, P P B BOH] OH K b = + \u2212 log [ ] [ or, ( ) log / / 14 9 5 4 5 \u2212 = + P K b a a \u2234 = \u21d2 = \u00d7 \u2212 P K K b b 5 6 2 5 10 6 . . At equivalent point: P P C) H K b => \u2212 + 1 2 ( log or, 4 5 1 2 5 6 . ( . log => \u2212 + \u21d2 = C) C 0.25 M Now, n n HCl used B formed = + or, V 0.5 V V ml HCl \u00d7 = + \u00d7 \u21d2 = 1000 100 0 25 1000 100 ( ) .\n7.49 Ionic Equilibrium HINTS AND EXPLANATIONS Finally, n n BOH takes Hcl used for equivalent point = or gm , . . w w 45 100 0 5 1000 2 25 = \u00d7 \u21d2 = \u2234 Percentage purity of base = \u00d7 = 2 25 2 5 100 90 . . %
Answer: A, C, D
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Solution: CO H O HCO OH K Kw K M M M h a 1 2 3 2 2 3 0 5 2 1 \u2212 \u2212 \u2212 \u2212 \u2212 + + + = = \u00d7 ( . ) ( ) ( ) ; x x y x y \u001f \u21c0 \u001f \u21bd \u001f \u001f 0 0 4 \u2212 HCO H O H CO OH K Kw K M M M 2 h a 3 2 3 9 2 1 2 5 10 \u2212 \u2212 \u2212 \u2212 + + + = = \u00d7 ( ) ( ) ; . x y y x y \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 2 10 0 5 0 5 10 4 2 \u00d7 = \u2212 \u22c5 + \u2212 \u2248 \u22c5 \u21d2 = \u2212 \u2212 ( ) ( ) ( . ) . x y x y x x x x and 2 5 10 2 5 10 9 9 . ( ) ( ) . \u00d7 = \u22c5 + \u2212 \u2248 \u22c5 \u21d2 = \u00d7 \u2212 \u2212 y x y x y y x x y Now, h x = = 0 5 0 02 . . P P OH H = \u2212 + \u2248 \u2212 = \u21d2 = \u2212 log( ) log( ) . . x y 10 2 0 12 0 2 and [H 2 CO 3 ] = y = 2.5 \u00d7 10 \u20139 M
Answer: A, B, D
\nOriginal PDF solution page
Open page 386 in PDFExtracted text
Solution: N H CH COOH N H CH COO NH CH P P K a 1 K a2 + = + \u2212 = \u23af \u2192 \u23af\u23af\u23af\u23af \u23af \u2192 \u23af\u23af\u23af\u23af 3 2 2 22 3 2 9 78 3 2 . . C COO \u2212 P P P H K K = + = + = 1 2 1 2 2 22 9 78 6 0 1 2 ( ) ( . . ) . a a Now, K a 1 3 3 2 22 6 3 10 0 01 10 = \u21d2 = \u00d7 \u2295 \u2212 + \u2295 \u2212 \u2212 \u2295 [NH ][H ] [NH ] . [NH . CH COO CH COOH C 2 2 H H COOH 2 ] \u2234 = = \u00d7 \u2295 \u2212 \u2212 [NH ] . . 3 5 78 6 10 1 7 10 CH COOH M 2 % of glycine in cationic form = \u00d7 \u00d7 = \u2212 1 7 10 0 01 100 0 017 6 . . . %
Answer: C
\nOriginal PDF solution page
Open page 386 in PDFExtracted text
Solution: At equivalent point, the solution should be acidic.
Answer: C
\nOriginal PDF solution page
Open page 386 in PDFExtracted text
Solution: Sodium acetate solution is basic.
Answer: C
\nOriginal PDF solution page
Open page 386 in PDFExtracted text
Solution: For precipitation of Fe(OH) 2 , [ ] . . . min max min OH P P OH H \u2212 \u2212 \u2212 = \u00d7 = \u00d7 \u21d2 = \u21d2 = 8 10 0 02 2 10 6 7 7 3 16 7 For precipitation of Fe(OH) 3 , [ ] . . . min / max min OH P P OH H \u2212 \u2212 \u2212 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u21d2 = \u21d2 = 4 10 0 05 2 10 8 7 5 3 28 1 3 9
Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 386 in PDFExtracted text
Solution: K x x x = \u00d7 = = \u2212 \u21d2 = 1.5 10 Dimer] [Monomer] 2 2 2 0 1 2 5 120 [ ( . ) \u2234 [ ] [ . Dimer Monomer] = \u2212 = x x 0 1 2 5 2
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 386 in PDFExtracted text
Solution: K x x x = \u00d7 = = \u2212 \u21d2 = \u00d7 \u2212 \u2212 3.6 Dimer Monomer 10 0 1 2 3 6 10 2 2 4 ( ) [ ] ( . ) . \u2234 (D ) [Monomer] . . imer = \u2212 \u21d2 = x x x 0 1 2 0 1 9 2500
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 386 in PDFExtracted text
Solution: [ ] . H M + \u2212 \u2212 \u2248 \u00d7 \u00d7 = \u00d7 0 1 2 10 2 10 5 3 (Dimerization is negative as Q. 2) \u2234 P H = 2.85\n7.50 Chapter 7 HINTS AND EXPLANATIONS Comprehension II
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: [ ] . CH COOH M O 3 3 3 8 0 7 10 10 10 7 10 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 The solution is so dilute that we may assume almost complete dissociation of acid. \u2234 \u2248 \u00d7 + \u2212 [ ] H M acid 7 10 8 Now, H O H OH M M 2 7 10 8 \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u00d7 + \u2212 \u2212 + ( ) x x 10 7 10 7 10 14 8 8 \u2212 \u2212 \u2212 = \u00d7 + \u22c5 \u21d2 = \u00d7 ( ) x x x \u2234 = \u2212 \u00d7 + = \u2212 P H log( ) . 7 10 6 85 8 x
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: CH COOH CH COO H Eqn M 7 10 M M 3 3 8 7 10 8 14 10 8 . y x \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2212 + \u00d7 \u00d7 \u2212 + = \u00d7 \u2212 \u2212 + Now, 2 0 10 7 10 14 10 5 8 8 . \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 y \u2234 y = 4.9 \u00d7 10 \u201310 M Comprehension III
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: K a = \u00d7 = \u00d7 \u2212 \u2212 ( ) . . 8 10 0 2 3 2 10 3 2 4
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: K a = \u00d7 = \u00d7 \u00d7 \u2212 + 3 2 10 1 0 0 8 0 2 4 . [ ] ( . . ) . H \u2234 [H + ] = 8 \u00d7 10 \u20135 = \u21d2 P H = 4.1 Comprehension IV
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: p p NH NH NH H Ka O = + = + = + + ( ) log [ ] [ ] . log . . . 4 3 4 9 3 0 8 0 2 9 9
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: NH OH H NH H O M M M M Final M 4 0 8 0 3 0 4 0 2 0 5 2 0 5 . . . . . + + + \u2248 + \u001f \u21c0 \u001f \u21bd \u001f \u001f p p NH NH H K O O a = + = + = + + (NH ) log [ ] [ ] . log . . . 4 3 4 9 3 0 5 0 5 9 3
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: H + added in excess. Final [H + ] = 1.0 \u2013 0.8 = 0.2 M \u2234 P H = \u2013 log(0.2) = 0.7 Comprehension V
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: No hydrolysis \u21d2 p H = 7.0
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: Concentration of KAl(SO H O M 4 2 2 12 11 85 474 100 1000 0 25 ) . . / / . = = Al H O Al(OH H M M M 3 0 25 2 2 + \u2212 + + + + ( . ) ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 1 4 10 0 25 0 25 1 87 10 5 2 3 . ( . ) . . ] \u00d7 = \u22c5 \u2212 \u21d2 = \u00d7 \u2212 \u2212 + x x x x x \u001b M = [H
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: SO H O HSO OH M 4 2 0 5 2 4 \u2212 \u2212 \u2212 \u2212 + + ( . ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 10 1 25 10 0 5 0 5 6 32 10 19 2 2 7 \u2212 \u2212 \u2212 \u00d7 = \u22c5 \u2212 \u21d2 = \u00d7 . ( . ) . . x x x x x \u001b M \u2234 = \u00d7 = \u00d7 + \u2212 \u2212 \u2212 [ ] . . H M 10 6 32 10 1 58 10 14 7 8
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 387 in PDFExtracted text
Solution: 1 4 10 1 25 10 0 25 0 5 0 25 0 5 5 2 2 . . ( . )( . ) . . \u00d7 \u00d7 = \u22c5 \u2212 \u2212 \u00d7 \u2212 \u2212 x x x x x \u001b \u2234 x = 1.18 \u00d7 10 \u20132 Now, 1 4 10 1 18 10 0 25 5 2 . . [ ] . \u00d7 = \u00d7 \u00d7 \u2212 \u2212 + H \u2234 [H + ] = 2.97 \u00d7 10 \u20134 M Al SO H O Al(OH HSO M M M 3 0 25 0 25 4 2 0 5 2 2 0 4 + \u2212 \u2212 + \u2212 + + + . Eqn. ( . ) . ) x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 0 0 x (0.5 \u2013 x )M\n7.51 Ionic Equilibrium HINTS AND EXPLANATIONS Comprehension VI
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: PuO H O PuO (OH H 2 2 0 01 0 01 2 2 1 6 10 4 + \u2212 \u2248 + + = \u00d7 + + \u2212 . . . ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2234 = \u22c5 \u2212 = \u00d7 \u2212 K x x x x h 0 01 0 1 2 56 10 2 6 . . . \u001b
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: K Kw K K a b b = \u21d2 = \u00d7 \u2212 3 9 10 9 . Comprehension VII
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: p P P H K K a 2 a = + = + = 1 2 1 2 8 13 1 0 5 3 ( ) ( ) .
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: For 2nd equivalent point, n n NaoH H PO = \u00d7 2 3 4 V V 40 ml \u00d7 = \u00d7 \u00d7 \u21d2 = 0 5 1000 2 100 0 1 1000 . . After adding HCl, HPO H H O millimole Final 5 millimole 0 4 2 10 5 2 4 0 5 \u2212 + \u2248 \u2212 + \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2234 = + = P H 8 5 5 8 0 log .
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: S K OH M sp = = \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 [ ] . ( ) . 2 30 6 2 18 4 0 10 10 4 0 10 Comprehension VIII
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: P P [HCO H CO H K O O a = + = + = \u2212 log ] [ ] . log . 3 2 3 6 4 8 1 7 3
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: 7 4 6 4 1 10 3 2 3 2 3 3 . . log [ ] [ ] [ ] [ ] = + \u21d2 = \u2212 \u2212 HCO H CO H CO HCO O O O O
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: More CO 2 should dissolve in solution. Comprehension IX
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: pH P C)=7+ a 2 = + + + = 7 1 2 1 2 10 6 1 12 3 ( log ( . log ) . K
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: CO H HCO M Final M M M 3 2 50 1 75 25 75 25 1 75 3 0 25 75 0 \u2212 \u00d7 + \u00d7 \u2212 + \u2248 \u001f \u21c0 \u001f \u21bd \u001f \u001f pH CO HCO a 2 O O = + = + = \u2212 \u2212 P K log [ ] [ ] . log / / . 3 2 3 10 6 1 3 1 3 10 6
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: CO H HCO M Final M M 3 2 50 1 100 0 50 1 100 3 0 50 100 0 \u2212 \u00d7 \u2248 + \u00d7 \u2212 + \u2248 \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2234 = + = + = pH a 1 a 1 2 1 2 5 4 10 6 8 0 2 ( ) ( . . ) . P P K K
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: CO H HCO M Final M M M 3 2 50 1 125 0 75 1 125 25 125 3 0 50 125 \u2212 \u00d7 + \u00d7 \u2212 + \u001f \u21c0 \u001f \u21bd \u001f \u001f HCO H H CO M Final M M M 3 50 125 25 125 25 125 0 2 3 0 25 125 \u2212 + = + \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2234 = + = + = \u2212 pH [HCO H CO a O O P K log ] [ ] . log / / . 3 2 3 5 4 25 125 25 125 5 4
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 388 in PDFExtracted text
Solution: [H CO ] M 2 3 = \u00d7 = 50 10 150 1 3 \u2234 = + = + = pH C a 1 1 2 1 2 5 4 1 3 2 94 ( log ) ( . log ) . P K\n7.52 Chapter 7 HINTS AND EXPLANATIONS Comprehension X
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: N H CH COOH N H CH 3 3 K K 3 3 b + = \u00d7 = \u00d7 + \u2212 \u2212 \u23af \u2192 \u23af\u23af\u23af\u23af\u23af \u2190 \u23af \u23af\u23af\u23af\u23af\u23af \u2212 \u2212 \u2212 2 12 1 3 2 5 10 4 10 . a \u2212 \u2212 \u23af \u2192 \u23af\u23af\u23af\u23af\u23af \u2190 \u23af \u23af\u23af\u23af\u23af\u23af \u2212 \u2212 \u2212 = \u00d7 = \u00d7 \u2212 \u2212 \u2212 COO N H CH COO K K a b 2 10 1 5 1 6 10 6 25 10 2 2 . . Required K K b b = = \u00d7 \u00d7 \u2212 1 6 25 10 10 5 .
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: \u2234 = + = + = pH a 1 a 1 2 1 2 2 4 9 8 6 1 2 ( ) ( . . ) . P P K K Comprehension XI
Comprehension 10 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: Moles of Cu reacted = 6 35 10 63 5 10 3 4 . . \u00d7 = \u2212 \u2212
Comprehension 11 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: ln . K G RT K eq eq = \u0394 \u2212 = \u2212 \u00d7 \u2212 \u00d7 = \u21d2 >>> \u00b0 120 10 8 0 300 50 1 3 \u2234 = + [ ] Ag 2 \u00d7 Mole of Cu reacted = 2 \u00d7 10 \u20134 M
Comprehension 11 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: K sp Ag M = = \u00d7 = \u00d7 + \u2212 \u2212 \u2212 [ ][BRO ] ( ) 3 4 2 8 2 2 10 4 10 Comprehension XII
Comprehension 11 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: S K AgCN CN M sp = = \u00d7 = \u00d7 \u2212 \u2212 \u2212 [ ] [ ] . . . 1 0 10 0 02 5 0 10 16 15
Comprehension 12 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: AgCN(s) CN Ag(CN S M SM eq sp + = \u00d7 = \u2212 \u2212 \u2212 ( . ) ) 0 02 2 15 \u001f \u21c0 \u001f \u21bd \u001f \u001f K K K f 15 0 02 0 3 16 1 875 10 2 = \u2212 \u21d2 = = \u00d7 \u2212 s s s M . . .
Comprehension 12 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: s [Ag ] [Ag(Cu Cu Cu M sp sp = + + \u22c5 \u22c5 + \u2212 \u2212 \u2212 ) ]; [ ] [ ] 2 5 K K K f For minimum solubility: ds d[Cu ] \u2212 = 0 or, \u2212 + \u22c5 = \u21d2 = = \u00d7 \u2212 \u2212 \u2212 K K K f sp sp f Cu Cu K M [ ] [ ] . 2 9 0 1 2 58 10
Answer: D
\nOriginal PDF solution page
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Solution: For acidic solution, pH < 7.0 at 25\u00b0 C.
Answer: A
\nOriginal PDF solution page
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Solution: If there were no common ion effect, P H should lie in between 7.0 and 7.3
Answer: D
\nOriginal PDF solution page
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Solution: [ ] ] . H M [H M HCl HCOOH + \u2212 + \u2212 = < = \u00d7 10 3 16 10 4 2
Answer: D
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: Dilution results decrease in concentration of BOH (aq), B + (aq) as well as OH \u2013 (aq).
Answer: D
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: The pH of buffer remains constant on slight dilution but for acidic solution, the dilution results in the decrease in [H + ] and hence, increase in pH.
Answer: C
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: pH of buff er containing H A and A \u2013 may be less than, greater than or equal to 7.0.
Answer: C
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Solution: K K a b ( ( ) CH COOH) NH OH 3 = 4 and hence, CH 3 COONH 4 solution is also neutral. But, it undergoes hydrolysis.
Answer: A
\nOriginal PDF solution page
Open page 389 in PDFExtracted text
Answer: C
\nOriginal PDF solution page
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Solution: Reaction occurs but at equivalent point, pH will be less than
Answer: D
\nOriginal PDF solution page
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Solution: 10. As dilution does not change the concentration of ions in saturated solution, the mole of ions will increase.\n7.53 Ionic Equilibrium HINTS AND EXPLANATIONS
Answer: A \u2192 P, Q; B \u2192 Q, R; C \u2192 R, S; D \u2192 T
\nOriginal PDF solution page
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Solution: True electrolytes produce ions in pure liquid form as well as in solution. Potential electrolytes are molecular in pure liquid state but it produces ions in solution.
Answer: A \u2192 Q, S; B \u2192 T; C \u2192 P; D \u2192 R
\nOriginal PDF solution page
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Solution: (P) [OH\u2013] to just start precipitation = K sp Mg [ ] 2 + = \u00d7 \u00d7 = \u2212 \u2212 \u2212 2 10 2 10 10 6 3 1 5 . \u2234 p OH = 1.5 \u21d2 P H = 12.5 (Q) [ ] [ ] . max / / OH Al M sp \u2212 + \u2212 \u2212 = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f = K 3 1 3 28 1 3 9 10 0 1 10 \u2234 = \u21d2 = p p OH H min max . 9 5 0 (R) CH COOH CH COO 3 M = M 3 M = 0 1 0 1 10 11 0 1 11 0 1 10 11 1 1 . . . . \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 \u001f \u21c0 \u001f \u21bd \u001f \u001f 1 1 M H + + ? Now, K a = = \u00d7 \u21d2 = \u2212 + + \u2212 10 1 11 0 1 11 10 5 6 [H ] . [ ] H M \u2234 p H = 6.0 (S) [ ] . . H C M pH a + \u2212 \u2212 = \u22c5 = \u00d7 = \u21d2 = K 10 0 001 10 5 0 7 5 (T) A H O HA OH M M M \u2212 \u00d7 \u2212 \u2212 \u2212 \u2212 + + = \u00d7 ( ) ; 6 10 2 6 5 2 10 x x x a Kw K \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 10 6 10 1 10 6 5 5 \u00d7 = \u22c5 \u00d7 \u2212 \u21d2 = \u00d7 \u2212 \u2212 \u2212 x x x x ( ) \u2234 P OH = 5 \u21d2 pH = 9
Answer: A \u2192 R; B \u2192 Q, T; C \u2192 P, S; D \u2192 P, Q, R
\nOriginal PDF solution page
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Answer: A \u2192 R; B \u2192 Q; C \u2192 P; D \u2192 S; E \u2192 T
\nOriginal PDF solution page
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Solution: (A) pH [H A H A] a 2 O O = + = \u2212 P K 1 3 4 0 log ] [ . (B) pH [HA H A ] a O O = + = \u2212 \u2212 P K 2 2 2 8 0 log ] [ . (C) pH [A HA ] a O O = + = \u2212 \u2212 P K 3 3 2 12 0 log ] [ . (D) pH a a 2 = + = + = 1 2 1 2 4 8 6 0 1 ( ) ( ) . P P K K (E) pH K K a a = + = + = 1 2 1 2 8 12 1 0 0 2 3 ( ) ( ) . P P
Answer: A \u2192 Q, R; B \u2192 P; C \u2192 P, S
\nOriginal PDF solution page
Open page 390 in PDFExtracted text
Solution: (A) K K a b ( ( [ ] . H O) H O)= Kw H O 2 2 2 14 16 10 1000 18 1 8 10 = = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u2212 (C) On increasing temperature, Kw increases and hence, P Kw decreases.
Answer: 8
\nOriginal PDF solution page
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Solution: [ ] log[ ] D K M P D W D + \u2212 \u2212 + = = = \u21d2 = \u2212 = 10 10 8 16 8
Answer: 4
\nOriginal PDF solution page
Open page 390 in PDFExtracted text
Solution: [ . HCOOH] M C O = \u00d7 = = 1 15 10 46 25 3 HCOOH HCOOH HCOOH HCOO C C \u2212 \u2212 + \u2212 + + x x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 From given data: x = = \u2212 K M 10 3 \u2234 Percentage of HCOOH molecules converted into HCOO \u2013 = \u00d7 = \u00d7 = \u00d7 \u2212 \u2212 x C 100 10 25 100 4 10 3 3
Answer: 1
\nOriginal PDF solution page
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Solution: [ ] / / . NH M O 3 3 10 17 100 0 85 10 5 = \u00d7 =\n7.54 Chapter 7 HINTS AND EXPLANATIONS [ ] ) OH C Kw K (NH C M a \u2212 + \u2212 \u2212 \u2212 = \u22c5 = \u22c5 = \u00d7 \u00d7 = K b 4 14 10 2 10 5 10 5 10 \u2234 = = = + \u2212 \u2212 \u2212 \u2212 [ ] [ ] H O Kw OH M 3 14 2 12 10 10 10
Answer: 6
\nOriginal PDF solution page
Open page 391 in PDFExtracted text
Solution: [ ] . H K C M a + \u2212 \u2212 = \u22c5 = \u00d7 \u00d7 = 4 10 0 0025 10 10 6 \u2234 = \u2212 = \u2212 P H log10 6 6
Answer: 2
\nOriginal PDF solution page
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Solution: [ . . H SO ] M 2 3 O = = 1 28 64 0 02 H SO H HSO 2 3 M M M ( . ) 0 02 3 \u2212 + \u2212 + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, K x x x x x a M pH = = \u22c5 \u2212 \u21d2 = \u21d2 = \u2212 = \u2212 10 0 02 0 01 2 2 ( . ) . log
Answer: 3
\nOriginal PDF solution page
Open page 391 in PDFExtracted text
Solution: pH HC H O [H C H O a 4 4 O O = + = + = \u2212 P K 1 6 2 4 4 6 3 3 18 8 18 8 30 150 3 log [ ] ] . log . / . /
Answer: 5
\nOriginal PDF solution page
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Solution: Buff er capacity, \u03b2 = \u2212 \u0394 = \u2212 \u2212 = + [ ] . / . ( . ) H added H P 0 05 0 2 0 05 5
Answer: 5
\nOriginal PDF solution page
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Solution: HA OH A millimole 0 millimole Equ.point a \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 \u2212 \u00d7 36 12 0 1 0 0 3 . . . .612 2 millimole H O + A H HA mmole Final mmole mmole \u2212 + + \u00d7 3 612 1 806 18 06 0 1 0 0 1 . . . . . \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f 8 806 mmole \u2234 = + = + = \u2212 pH A HA] a O O P K log [ ] [ log . . 5 1 806 1 806 5
Answer: 6
\nOriginal PDF solution page
Open page 391 in PDFExtracted text
Solution: \u2234 = + = + = pH a a 2 1 2 1 2 2 28 9 72 6 1 ( ) ( . . ) P P K K
Answer: 4
\nOriginal PDF solution page
Open page 391 in PDFExtracted text
Solution: For appearance of only ln + colour, log [ln ] [ln . . . log + = \u2212 = = OH] 4 6 3 4 2 0 6 4 \u2234 = + [ln ] [ln OH] 4 Four-digit Integer Type
Answer: 0160
\nOriginal PDF solution page
Open page 391 in PDFExtracted text
Solution: C H NH H O C H NH OH M Eqn. M M 2 M M 6 5 2 0 2 0 2 0 2 6 5 3 0 10 8 . ( . ) . \u2212 + = + + \u2212 x x \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2212 \u2212 + CM (C ) M CM x \u001b Now, K C b = \u21d2 \u00d7 = \u00d7 + \u2212 \u2212 \u2212 [C H NH ][OH ] C H NH . 6 5 3 6 5 2 10 8 4 10 10 0 2 \u2234 C = 8 \u00d7 10 \u20133 M Now, mass of NaOH added = \u00d7 \u00d7 \u00d7 = = \u2212 8 10 1000 500 40 0 16 160 3 . gm mg
Answer: 0369
\nOriginal PDF solution page
Open page 391 in PDFExtracted text
Solution: K a = \u21d2 \u00d7 = \u00d7 \u00d7 + \u2212 \u2212 \u2212 \u2212 [ ][CH COO ] [CH COOH . [CH COO ] . H ] 3 3 5 4 3 1 8 10 4 10 0 2 \u2234 = \u00d7 \u2212 \u2212 [CH COO ] 3 3 9 10 M \u2234 Mass of CH 3 COONa added 9 10 500 1000 82 0 369 3 \u00d7 \u00d7 \u00d7 = = \u2212 . gm 369 mg
Answer: 1100
\nOriginal PDF solution page
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Solution: H SO H HSO 2 3 CM Final 0 CM CM \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + 0 4 0 H A OH HA mmole Final 0 mmole mmole 2 20 0 09 30 0 06 0 1 8 + \u2212 \u00d7 \u00d7 + + . . . \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f H H O 2\n7.55 Ionic Equilibrium HINTS AND EXPLANATIONS HSO H SO Fianl C )M C M =0.01 M M 4 4 2 \u2212 \u2212 + + \u2212 + ( ( ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 1 2 10 0 01 6 11 2 . . \u00d7 = \u00d7 \u2212 \u21d2 = \u2212 x x x C C and C + x = 0.01 \u21d2 C M = \u00d7 0 01 11 17 .
Answer: 0080
\nOriginal PDF solution page
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Solution: P P NaHCo [H Co H K O O a = + log [ ] ] 3 2 3 or, 7 4 6 1 10 2 80 . . log = + \u00d7 \u00d7 \u21d2 = V 5 V ml
Answer: 0060
\nOriginal PDF solution page
Open page 392 in PDFExtracted text
Solution: [ ] . / / . HSO M O 4 1 8 120 100 1000 0 15 \u2212 = = HSO H SO M M M 4 0 15 4 2 \u2212 \u2212 + \u2212 + ( . ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 4 10 0 15 6 10 60 2 2 \u00d7 = \u22c5 \u2212 \u21d2 = \u00d7 = \u2212 \u2212 x x x x . M millimole/L
Answer: 5340
\nOriginal PDF solution page
Open page 392 in PDFExtracted text
Solution: Al H O H O Al H O H O C M Now, 10 OH M M ( ) ( ) ( ) 2 6 3 2 2 5 3 1 5 2 + \u2212 = + = \u2212 + + + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 0 0 3 \u2212 \u2248 M C 0.1M \u2234 Mass of Al(OH) 3 added = 400 \u00d7 0.1 \u00d7 133.5 = 5.34 gm = 5340 mg
Answer: 0050
\nOriginal PDF solution page
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Solution: 9 18 60 40 . log = + P K a (1) 9 00 100 . log = + \u2212 P x x K a (2) \u2234 x = 50
Answer: 0960
\nOriginal PDF solution page
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Solution: I I I M Eqn. (0.05 M M 2 12 7 254 0 05 0 1 0 1 3 0 . . ) . . = \u2212 \u2212 \u2212 \u2212 + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f = = 0 254 254 0 001 . . \u2234 x = 0.049 Now, K x x x c = \u2212 \u2212 = \u00d7 = ( . )( . ) . . . 0 05 0 1 0 049 0 001 0 051 960
Answer: 0740
\nOriginal PDF solution page
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Solution: Concentration of Ca(OH) 2 in its saturated solution = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 K sp M 4 3 2 10 4 0 02 1 3 5 1 3 / / . . Now, Ca OH Ca(OH) s M M 2 0 02 0 04 2 2 + \u2212 + . . ( ) \u001f \u21c0 \u001f \u21bd \u001f \u001f 0.01 M (0.02 + 0.8) M = 0.82 M Equ (0.01 \u2013 x )M = 0 (0.82 \u2013 2 x ) M = 0.8M \u2234 = \u00d7 = \u00d7 << + \u2212 \u2212 [ ] . ( . ) . Ca left 2 5 2 5 3 2 10 0 8 5 10 0 01 \u2234 Ca(OH) 2 precipitated = 0.01 mole = 0.01 \u00d7 74 = 0.74 gm = 740 mg
Answer: 5500
\nOriginal PDF solution page
Open page 392 in PDFExtracted text
Solution: S M = = \u2212 0 0055 550 100 1000 10 4 . / / \u2234 K sp of Ca(pam) 2 = 4S 3 = 4 \u00d7 10 \u201312 M Now, Ca pam Ca(pam) M Final =0 M 0.1M 2 40 40 10 10 10 0 1 2 6 2 3 2 + \u00d7 = \u2212 \u2212 + / . ( \u001f \u21c0 \u001f \u21bd \u001f \u001f s s) \u2234 Ca(pam) 2 participated = 10 \u20133 \u00d7 10 = 0.01 mole = 0.01 \u00d7 550 = 5.50 gm = 550 mg On adding NaOH
![\"Liquid](\"./assets/chapter-packs/liquid-solution/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 517 in PDFExtracted text
Solution: P K X N H N 2 2 = \u22c5 ( ) Solution or 5 0 8 1 0 10 10 10 10 5 5 2 2 2 \u00d7 = \u00d7 \u00d7 + \u00d7 . ( . ) n n n N N N \u001f \u2234 n N 2 4 10 4 = \u00d7 \u2212
Answer: C
\nOriginal PDF solution page
Open page 517 in PDFExtracted text
Solution: p K X K n n m P V H H = \u22c5 \u2248 \u22c5 \u21d2 \u22c5 gas liq gas liq \u03b1 \u2234 = \u22c5 \u22c5 \u21d2 = \u00d7 \u00d7 \u21d2 = m m P V P V m m m 2 1 2 2 1 1 2 2 5 2 1 1 10 m Now, P K n n K n P V RT H H = \u22c5 = \u22c5 \u22c5 gas liq liq \u2234 Volume of gas dissolved, V RT K n H = \u22c5 liq (Volume of gas dissolved is independent of pressure of gas) \u2234 = \u21d2 = \u21d2 = V V V V V V V 2 1 2 1 2 2 2 1 2 , , liquid liquid ml V ml\n10.31 Liquid Solution HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: P X P Hg Hg = \u22c5 = \u00d7 \u00d7 + \u00d7 = \u00d7 \u2212 \u2212 \u2212 total t 0 8 10 200 0 8 10 200 50 4 28 720 1 6 10 3 3 3 . . . . o orr
Answer: C
\nOriginal PDF solution page
Open page 518 in PDFExtracted text
Solution: Final vapour pressure and hence, the composition of both solutions must be same. As solution in beaker (A) has higher concentration, its vapour pressure is low. Hence, water from (B) will transfer in (A) as vapour. \u2234 + = \u2212 \u21d2 = 20 200 10 100 33 33 x x x .
Answer: B
\nOriginal PDF solution page
Open page 518 in PDFExtracted text
Solution: Y P P X P P A A A A = = \u22c5 \u00b0 total total 1 1 1 1 X Y P Y P Y P Y P P P P P A A A A A A B A A B B A B = \u22c5 \u00b0 \u22c5 \u00b0 + \u2212 \u00b0 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u00b0 \u00b0 + \u00b0 \u2212 \u00b0 \u00b0 ( )
Answer: C
\nOriginal PDF solution page
Open page 518 in PDFExtracted text
Solution: n n n n P P P Q P Q P Q n \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u00b0 \u00b0 \u239b \u239d \u239c \u239e \u23a0 \u239f 2 nd condense initial Where n = number of condensation steps. Now, n n P Q \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = final 1 1 300 100 9 1 2 \u2234 = + = X P 9 9 1 0 90 .
Answer: B
\nOriginal PDF solution page
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Solution: The mole fraction of A in distillate, \u2032 = = \u22c5 \u00b0 = \u00d7 \u00d7 + \u00d7 = X Y X P P A A A A total 1 4 100 1 4 100 3 4 80 5 17 Now, V.P. of distillate, P X P X P A A B B = \u2032 \u22c5 \u00b0 + \u2032 \u22c5 \u00b0 = \u00d7 + \u00d7 = 5 17 100 12 17 80 85 88 . m m Kg
Answer: D
\nOriginal PDF solution page
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Solution: Let the final composition: liquid (10 mole): A = x mole, B = (10 \u2013 x ) mole Vapour (10 mole): A = (10 \u2013 x ) mole, B = x mole Now, Y Y X X P P x x x x A B A B A B = \u22c5 \u00b0 \u00b0 \u21d2 \u2212 = \u2212 \u22c5 10 10 200 100 \u2234 x = 4.14 Now, p X P X P x x A A B B = \u22c5 \u00b0 + \u22c5 \u00b0 = \u00d7 + \u2212 \u00d7 10 200 10 10 100 = 141.4 torr
Answer: A
\nOriginal PDF solution page
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Solution: 1 0 4 0 4 0 6 1 2 3 2 2 3 P Y P Y P P A A B B total total atm = \u00b0 + \u00b0 = + = \u21d2 = . . . .
Answer: A
\nOriginal PDF solution page
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Solution: If the solution were ideal, P total mm kg = \u00d7 + \u00d7 = 10 30 90 20 30 87 88 As the solution of phenol and aniline shows negative deviation, the V.P. must be less than 88 mm kg.
Answer: C
\nOriginal PDF solution page
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Solution: For ideal behavior, P total = 0.25 \u00d7 512 + 0.725 \u00d7 344 = 386 mm Hg < 600 mm Hg Hence, the solution shows positive deviation \u21d2 \u0394 H mix = positive
Answer: C
\nOriginal PDF solution page
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Solution: For increase in temperature, the solution shows negative deviation ( \u0394 H = negative).
Answer: C
\nOriginal PDF solution page
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Solution: n n P P Chlorobenzene water Chlorobenzene water = \u00b0 \u00b0 or, x x x / . ( ) / . . . 112 5 100 18 9 031 10 7 031 10 7 031 10 64 4 4 4 \u2212 = \u00d7 \u2212 \u00d7 \u00d7 \u21d2 =
Answer: D
\nOriginal PDF solution page
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Solution: As the available surface area for solvent molecules decreases, the rate of vaporization decreases.
Answer: B
\nOriginal PDF solution page
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Solution: P P P n n m m m m \u00b0 \u2212 = \u21d2 = \u21d2 = 1 2 1 2 2 1 5 95 0 3 57 10 / / . M M
Answer: B
\nOriginal PDF solution page
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Solution: P P X P \u00b0 \u2212 = \u22c5 \u00b0 1 10 = 0.2 \u00d7 P \u00b0 (1) 20 = X 1 \u00d7 P \u00b0 (2) From (1) and (2): X 1 = 0.4 \u21d2 X 2 = X solvent = 0.6
Answer: C
\nOriginal PDF solution page
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Solution: P P P n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 3000 2985 2985 5 100 18 179 1 / / . M M
Answer: A
\nOriginal PDF solution page
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Solution: P P P n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 0 85 0 845 0 845 0 5 39 78 169 . . . . / / M M\n10.32 Chapter 10 HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: P X P = \u22c5 \u00b0 = + \u00d7 = 2 1000 18 1 1000 18 12 3 12 08 . . K Pa
Answer: A
\nOriginal PDF solution page
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Solution: P X P = \u22c5 \u00b0 2 2 8 90 18 30 90 18 . = + \u00d7 \u00b0 M P and 2 9 108 18 30 108 18 . = + \u22c5 \u00b0 M P \u2234 = M 23
Answer: A
\nOriginal PDF solution page
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Solution: P P X P \u00b0 \u2212 = \u22c5 \u00b0 = + \u00d7 = 1 1 1 1000 18 760 13 44 . m m kg
Answer: D
\nOriginal PDF solution page
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Solution: P P P n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 89 78 89 89 2 100 78 178 . / / M M Now, the number of C-atoms in each molecule = 178 94 4 100 12 14 \u00d7 = . and the number of H-atoms in each molecule = 178 5 6 100 1 10 \u00d7 = . \u2234 Hydrocarbon is C 14 H 10 .
Answer: B
\nOriginal PDF solution page
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Solution: Loss is mass of solvent, w 1 a ( P \u00b0 \u2013 P ) and gain is mass of absorbent, w 2 a P \u00b0 \u2234 = \u00b0 \u2212 \u00b0 = \u21d2 = + \u21d2 = w w P P P X 1 2 1 0 05 2 05 40 40 100 18 288 . . M M M
Answer: C
\nOriginal PDF solution page
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Solution: C C M M M M M 1 2 10 20 6 67 30 1 3 = \u21d2 + = + \u21d2 = A B A B A B M .
Answer: A
\nOriginal PDF solution page
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Solution: As solution have same concentration, mixing will not change the total molar concentration.
Answer: A
\nOriginal PDF solution page
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Solution: Isopiestic refers to the same pressure. Blood is isotonic with 0.9 % ( w / v ) NaCl solution. \u2234 Osmolarity = \u00d7 \u2248 0 9 58 5 100 1000 2 0 31 . / . / . M
Answer: B
\nOriginal PDF solution page
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Solution: \u03c0 = = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 = CRT atm 2 5 58 5 100 1000 2 0 0821 300 21 05 . / . / . .
Answer: C
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m b b 0 104 0 52 2 98 1000 104 . . / / M M
Answer: D
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: Clausius\u2013Clapeyron equation: dT dP RT H P = \u0394 = \u00d7 \u00d7 = 2 2 3 2 350 4 9 10 50 . ( ) . K/atm
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 = \u00d7 = T K m b b 0 52 72 180 0 208 . . K \u2234 B.P. of solution = 373.208 K
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 \u2212 = \u00d7 \u00d7 T K m b b ( . . ) . . / 354 11 353 23 2 53 1 8 90 1000 M \u2234 M = 57.5
Answer: C
\nOriginal PDF solution page
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Solution: If molality of solution is \u2018 m \u2019, then P P P n n m m \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 760 750 750 1000 18 20 27 / Now, \u0394 = \u22c5 = \u00d7 = T K m b b 0 52 20 27 0 385 . . K \u2234 B.P. of solution = 100.385\u00b0 C
Answer: A
\nOriginal PDF solution page
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Solution: K H b = \u0394 = \u00d7 = 0 002 0 002 320 80 2 56 2 2 . ( ) . ( ) . T Cal/gm K/m o vap Now, \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m x x b b 0 32 2 56 6 4 32 200 1000 8 . . . / / \u2234 Molecular formula of sulphur = S x = S 8
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 = \u00d7 + \u2248 \u00b0 T K X b x b , / . solute C 32 1 128 1 128 94 94 0 25 \u2234 B.P. of solution = 110.75 + 0.25 = 111\u00b0 C
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m m m b b 1 04 0 52 2 . . Now, P P P n n P P \u00b0 \u2212 = \u21d2 \u00b0 \u2212 = \u21d2 \u00b0 = 1 2 750 750 2 1000 18 777 / torr
Answer: C
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m f f 0 93 1 86 7 93 1000 150 5 . . / / . M M\n10.33 Liquid Solution HINTS AND EXPLANATIONS
Answer: D
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m f f 0 93 1 86 36 1 2 60 . . / . M M If the molecular formula is C x H 2 x O x , then 12 x + 2 x + 16 x = 60 \u21d2 x = 2 \u2234 Molecular formula = C 2 H 4 O 2
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m m m f f 15 1 86 8 06 . .
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 \u0394 = \u21d2 \u0394 = \u21d2 \u0394 = \u00b0 T T K K T T f b f b f f 0 78 1 86 0 52 2 79 . . . . C \u2234 F.P. of solution = \u20132.79\u00b0 C
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 + \u0394 = + \u22c5 T T K K m f b f b ( ) or 4.76 = (1.86 + 0.52) \u00d7 w w / / . 342 100 1000 68 4 \u21d2 = gm
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m w w f f 5 8 5 120 425 1000 30 . / / gm
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 \u0394 = \u21d2 \u0394 = \u21d2 \u0394 = \u00b0 T T K K T T f b f b b b 0 7 5 17 5 0 2 . . . C \u2234 B.P. of solution = 90.2\u00b0C
Answer: A
\nOriginal PDF solution page
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Solution: If the molarity of solution is m , then P P P n n m m \u00b0 \u2212 = \u21d2 = \u21d2 = 1 2 2 100 1000 78 10 39 / Now, \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m K K f f f f 1 3 10 39 5 07 . . K/m
Answer: C
\nOriginal PDF solution page
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Solution: K K T H T H H H T T K K f b f b f b b f = \u00b0 \u0394 \u00b0 \u0394 \u21d2 \u0394 \u0394 = \u00b0 \u00b0 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 2 2 2 / / fus vap fus vap = \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 280 350 2 5 5 6 2 7 2 . .
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m K K f f f f 2 0 0 25 8 . . K/m
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 T K m f f For cane sugar solution: 273 15 272 85 5 342 95 1000 . . \u2212 ( ) = \u22c5 K f / / (1) For glucose solution: 273 15 5 180 95 1000 . / / \u2212 ( ) = \u22c5 T K f f (2) From (1) and (2), T f = 272.58 K
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 T K m f f For AB 2 solution: 2.55 = 5 1 1 2 20 1000 . / ( ) / \u00d7 + x y (1) For AB 4 solution: 1.7 = 5 1 1 4 20 1000 . / ( ) / \u00d7 + x y (2) \u2234 Atomic mass of A = x = 50 Atomic mass of B = y = 25
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m m m f f 0 744 1 86 0 4 . . . Now, \u03c0 = CRT = 0.4 \u00d7 0.0821 \u00d7 300 = 9.852 atm
Answer: C
\nOriginal PDF solution page
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Solution: \u03d5 = \u0394 \u22c5 = \u00d7 = T K m f f 0 93 1 86 0 4 1 25 . . . .
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m m m f f 1 0 1 80 1 1 8 . . . P X P = \u22c5 \u00b0 = + \u00d7 = 2 1000 18 1 1 8 1000 18 24 24 24 . . mm Hg
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 T K m f f \u2234 = \u00d7 = \u00d7 0 2 100 1000 1000 . / / K x K x y f f and 0.25 \u2234 Mass of ice separated out = 100 \u2013 y = 20 gm
Answer: B
\nOriginal PDF solution page
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Solution: 2 KI(aq) HgI K HgI aq particles added 2 4 particles ( ) ( ) ( ) ( ) 4 2 3 + \u23af \u2192 \u23af As the number of ions in solution decreases, osmotic pressure decreases.
Answer: A
\nOriginal PDF solution page
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Solution: Osmolarity of both solution should be equal \u2234 \u00d7 = \u00d7 + \u21d2 = 0 1 2 0 1 1 2 0 5 . . ( ) . \u03b1 \u03b1
Answer: A
\nOriginal PDF solution page
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Solution: Osmolarity = 0.2 \u00d7 3 = 0.6 M
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 \u21d2 = T K m n n f f 3 72 1 86 1 0 2 . . . \u2234 From each particle, two ions should form.
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m K b b b 1 43 1 1 0 9 1 2 1 . . \u2234 K b = 2.6 K/m
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m n f f 1 1 1 \u03b1 \u2234 = \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u21d2 = 1 96 4 9 2 122 25 1000 1 1 2 1 0 78 . . / / . \u03b1 \u03b1\n10.34 Chapter 10 HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: HA H A M M =0.01M M ( . ) 0 1 \u2212 + + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2212 \u2234 Osmolarity = (0.1 \u2013 x ) + x + x = 0.11 M Now, \u03c0 = CRT = 0.11 RT
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 = \u00d7 \u00d7 + \u00d7 = \u00b0 T K m f f 1 86 0 1 2 0 025 2 0 465 . [ . . ] . C \u2234 F.P. of solution = \u2013 0.465\u00b0 C
Answer: A
\nOriginal PDF solution page
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Solution: The eff ective molality will be in the range of 0.2 to 0.3 and \u0394 T f = K b \u22c5 m will be in the range of 1.86 \u00d7 0.2 = 0.372\u00b0 C to 1.86 \u00d7 0.3 = 0.558\u00b0 C.
Answer: A
\nOriginal PDF solution page
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Solution: Let the mixture contain x mole KCl and y mole NaCl. Then x \u00d7 74.5 + y \u00d7 58.5 = 3.125 (1) and ( ) . . . x y T K f f + \u00d7 = \u0394 = = 2 0 186 1 86 0 1 (2) \u2234 x y = 1 3
Answer: B
\nOriginal PDF solution page
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Solution: 2 2 0 2 A A Initial conc. nM ( M M Equilibrium conc. n x x \u2212 ) \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, \u0394 = \u22c5 = \u22c5 \u2212 + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m K n x x b b b 2 \u2234 x n T K b b = \u2212 \u0394 2 2 Now, K A A x n x n T K T K n c b b b b = = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u2212 \u0394 \u0394 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f [ ] [ ] ( ) 2 2 2 2 2 2 = \u22c5 \u2212 \u0394 \u0394 \u2212 \u22c5 K n K T T n K b b b b b ( ) ( ) 2 2
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 \u0394 = \u22c5 \u22c5 = \u00d7 \u00d7 = T A T B K A m K B m m m f f f f ( ) ( ) , , . / . / 1 2 1 86 2 2 79 3 1 1
Answer: B
\nOriginal PDF solution page
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Solution: Colloidal solutions have low value of any colligative property than the true solution of same composition.
Answer: C
\nOriginal PDF solution page
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Solution: If the complex dissociates into n ions, then \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 \u21d2 \u2248 T K m n n f f 0 0054 1 86 0 001 3 . . [ . ]
Answer: A
\nOriginal PDF solution page
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Solution: XCl X Cl M M 3 3 3 3 (s) \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + S S P P P n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = \u00d7 \u2212 1 2 2 17 25 17 20 17 20 4 1000 18 4 04 10 . . . / . S M S
Answer: C
\nOriginal PDF solution page
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Solution: (a) \u0394 = \u22c5 = \u00d7 = \u00b0 T K m f f 1 86 1 1 86 . . C \u2234 F.P. of solution = \u2013 1.86\u00b0 C (b) \u0394 = \u22c5 = \u00d7 = \u00b0 T K m b b 0 52 1 0 52 . . C \u2234 B.P. of solution should be 100.52\u00b0 C. As the solute dissociates completely above 100.26\u00b0 C, its actual \u0394 = \u00d7 = \u00b0 T b 0 52 2 1 04 . . C and hence, B.P. = 101.04\u00b0 C. (c) \u0394 = \u22c5 \u21d2 = \u00d7 T K m m f f 7 44 1 86 1 2 . . / solvent (as solute dimerizes) \u2234 m Solvent left = 0.125 kg \u2234 Percentage of water separated as ice = (1 \u2013 0.125) \u00d7 100 = 87.5 % (d) \u0394 = \u22c5 \u21d2 = \u00d7 T K m m b b 2 08 0 52 2 . . solvent (Complete dissociation) \u2234 m Solvent left = 0.5 kg Percentage of water evaporated = (1 \u2013 0.5) \u00d7 100 = 50 %
Answer: D
\nOriginal PDF solution page
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Solution: At constant temperature, the vapour pressure may be changed by changing the composition.
Answer: A
\nOriginal PDF solution page
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Answer: A, C
\nOriginal PDF solution page
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Answer: B, C, D
\nOriginal PDF solution page
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Solution: P AB = + = 75 22 2 48 5 . torr P BC = + = 22 10 2 16 torr P AC = + = 75 10 2 42 5 . torr P ABC = + + = 75 22 10 2 35 67 . t orr\n10.35 Liquid Solution HINTS AND EXPLANATIONS
Answer: A, B, C, D
\nOriginal PDF solution page
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Answer: A, C, D
\nOriginal PDF solution page
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Solution: A, C, D shows negative deviation.
Answer: C
\nOriginal PDF solution page
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Solution: (a) Mass percent of A = 50 \u21d2 n A : n B = 1 : 2 \u21d2 Azeotrope Hence, vapour will have the same composition of liquid. (b) Mass percent of A > 50 \u21d2 n A : n B = 1 : 2 L L V 0.0 1.0 V \u00b0 A \u03c1 \u00b0 B \u03c1 mole-fraction of B \u03c1 2 3 In this case, the vapour must be more rich in A than liquid. (c) X B = = > 3 4 0 75 2 3 . \u21d2 Pure A cannot be obtained. (d) X B = = < 3 5 0 60 2 3 . \u21d2 Pure A cannot be obtained in traces.
Answer: B, C, D
\nOriginal PDF solution page
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Solution: (a) On changing the solvent, K f will change.
Answer: A, B
\nOriginal PDF solution page
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Solution: F. P. and V. P. will become lower for X .
Answer: B, C
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: B, C, D
\nOriginal PDF solution page
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Answer: A, B, C, D
\nOriginal PDF solution page
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Solution: (a) P P P n n n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 2 2 760 740 740 1 37 \u2234 Moles of water separated as ice = 200 \u2013 37 = 163 (b) \u0394 = \u22c5 = \u00d7 \u00d7 = \u00d7 T K m f f 2 0 1 37 18 1000 2000 37 18 . ( ) / K \u2234 = \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f T K 273 2000 37 18 (c) For original solution: \u0394 = \u22c5 = \u00d7 \u00d7 T K m f f 2 1 200 18 1000 ( ) / \u2234 F. P. = 0 10 18 \u2212 \u0394 = \u2212 \u00b0 T C f (d) For final solution: P P P X \u00b0 \u2212 \u00b0 = = + = 1 1 1 37 1 38
Answer: B, C, D
\nOriginal PDF solution page
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Answer: C, D
\nOriginal PDF solution page
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Solution: 0.0 1.0 \u00b0 T A B T \u00b0 T B
Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Y X X P P X P P X A A A A B A A B A \u2212 = \u22c5 \u00b0 \u00b0 + \u22c5 \u00b0 \u2212 \u00b0 \u2212 ( ) = \u00b0 \u2212 \u00b0 \u2212 \u00b0 \u2212 \u00b0 \u00b0 + \u22c5 \u00b0 \u2212 \u00b0 = X P P X P P P X P P f X A A B A B A B A A B A ( ) ( ) ( ) ( ) 2 For maximum ( ), ( ) Y X d Y X dX A A A A A \u2212 \u2212 = 0 \u2234 X P P P P P A A B B A B = \u00b0 \u22c5 \u00b0 \u2212 \u00b0 \u00b0 \u2212 \u00b0
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: P P X P P P P B A A B A B total = \u00b0 + \u00b0 \u2212 \u00b0 = \u00b0 \u22c5 \u00b0 \u22c5 ( )\n10.36 Chapter 10 HINTS AND EXPLANATIONS Comprehension II
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: 1 2 5 0 4 3 5 0 6 0 5 0 3 P Y P Y P P A A B B total total bar = \u00b0 + \u00b0 = + \u21d2 = > / . / . . . As the applied pressure is less than equilibrium pressure, the system must be 100 % vapour.
Comprehension 1 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: First drop of liquid will form at 0.5 bar.
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: P 1 V 1 = P 2 V 2 \u21d2 0.3 \u00d7 10 = 0.5 \u00d7 V 2 \u21d2 V 2 = 6.0 dm 3
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: P X P X P X X A A B B A A total = \u22c5 \u00b0 + \u22c5 \u00b0 \u21d2 = \u00d7 + \u2212 \u00d7 0 5 0 4 1 0 6 . . ( ) . \u2234 X A = 0.5
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: Liquid composition: A \u2248 2 mole, B \u2248 3 mole \u2234 P total bar = \u00d7 + \u00d7 = 2 5 0 4 3 5 0 6 0 52 . . .
Comprehension 2 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: Y X P P A A A = \u22c5 \u00b0 = \u00d7 = total 2 5 0 4 0 52 4 13 . .
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: P X P X P X X A A B B B A total = \u22c5 \u00b0 + \u22c5 \u00b0 \u21d2 = \u00d7 + \u2212 \u00d7 0 51 0 4 1 0 6 . . ( ) . \u2234 X A = 0 45 . Now, Y X P P A A A = \u22c5 \u00b0 = \u00d7 = total 0 45 0 4 0 51 6 17 . . . Let moles of A and B in liquid form is x and y , respectively. X x x y Y x x y A A = + = = \u2212 \u2212 + \u2212 = 0 45 2 2 3 6 17 . ( ) ( ) and \u2234 n A (liquid) = x = 12 11 and n A (vapour) = 2 \u2013 x = 10 11
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: Final total moles of liquid = 5 20 100 1 \u00d7 = and total moles of vapour = 5 \u2013 1 =
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: Let the liquid contain x mole A . P x x x total = \u00d7 + \u2212 \u00d7 = \u2212 1 0 4 1 1 0 6 0 6 0 2 . . . . (1) and Y X P P x x x A A A = \u22c5 \u00b0 \u21d2 \u2212 = \u00d7 \u2212 total 2 4 1 0 4 0 6 0 2 . . . (2) From (2): x = 0.48 \u2234 From (1): P total = 0.504 bar Comprehension III
Comprehension 3 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: \u0394 H mix = 0
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: \u0394 G mix, m = RT [ X 1 \u22c5 ln X 1 + X 2 \u22c5 ln X 2 ] = \u00d7 \u22c5 + \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 2 300 1 3 1 3 2 3 2 3 ln ln = \u2013380 cal/mol
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: \u0394 = \u2212 \u0394 = \u2212 \u2212 = S G T m m mix, mix, . / 380 300 3 8 3 cal K-mol and \u0394 S mix = 3 3 8 3 3 8 \u00d7 = . . / cal K Comprehension IV
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: P X P X P B B T T total mm Hg = \u22c5 \u00b0 + \u22c5 \u00b0 = \u00d7 + \u00d7 = 10 20 100 10 20 40 70
Comprehension 4 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: Y X P P A B B = \u22c5 \u00b0 = \u00d7 = = total 0 5 100 70 5 7 0 714 . .
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: The vapour will contain almost 10 moles of both 1 0 5 100 0 5 40 57 14 P Y P Y P P B B T T total total mm kg = \u00b0 + \u00b0 = + \u21d2 = . . .
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: X Y P P B B B = \u22c5 \u00b0 = \u00d7 = total 0 5 57 14 100 0 286 . . .
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 523 in PDFExtracted text
Solution: Final system contains 10 moles of liquid and 10 moles of vapour. Let the moles of benzene in liquid be x . P X P X P x x B B T T total = \u22c5 \u00b0 + \u22c5 \u00b0 = \u00d7 + \u2212 \u00d7 10 100 10 10 40 or, P total = 40 + 6 x (1) Y X P P x x x x B B B = \u22c5 \u00b0 \u21d2 \u2212 = \u00d7 + \u21d2 = total 10 10 10 100 40 6 3 87 . From Equation (1): P total = 63.25 mm kg\n10.37 Liquid Solution HINTS AND EXPLANATIONS Comprehension V A + B Residual solution A = x mole B = y mole Condensate ( n A + n B) moles = ( n A + n B) moles 1 4 A = ( n A \u2013 x ) mole B = ( n B \u2013 y ) mole = ( n A + n B) 3 4 From question: 700 = + \u00d7 \u00b0 + + \u00d7 \u00b0 n n n P n n n P A A B A B A B B (1) 600 = + \u00d7 \u00b0 + + \u00d7 \u00b0 x x y P y x y P A B (2) x y n n A B + = + 1 4 ( ) (3) x x y + = 0 3 . (4) n x n n A A B \u2212 + = 3 4 0 75 ( ) . (5)
Comprehension 5 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: n B : n A = 29.51
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: P A \u00b0 = 807 4 . mm
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: P B \u00b0 = 511 1 . mm Comprehension VI
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: P RT V A m \u00b0 = = \u00d7 \u00d7 \u00d7 = 0 08 300 100 1 25 3800 1000 760 60 . . torr
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: P RT V B m \u00b0 = = \u00d7 \u00d7 \u00d7 = 0 08 300 50 1 00 7600 1000 760 48 . . torr
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: 1 1 54 60 1 48 5 9 P Y P Y P Y Y Y A A B B A A A total = \u00b0 + \u00b0 \u21d2 = + \u2212 \u21d2 = Comprehension VII
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: \u0394 = \u22c5 = \u22c5 T K m K f f f 50 M \u2234 \u0394 \u0394 = = T A T B f f B B ( ) : ( ) M : : M 3 1
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: Average molar mass of solute in S 1 M(S M M M 1 2 3 2 3 11 5 ) = \u00d7 + \u00d7 + = A B A and average molar mass of solute in S 2 . M( S 2 ) = 3 2 2 3 9 5 M M A B A M + \u00d7 + = \u2234 \u0394 \u0394 = ( ) ( ) = T S T S M S M S f f ( ) : ( ) : : 1 2 2 1 9 11 Comprehension VIII
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: \u0394 = \u22c5 = \u00d7 \u00d7 \u00d7 = T K m f f 2 0 0 1 0 9 46 1000 4 8 . . . . K \u2234 Freezing point of solution = 155.7 \u2013 4.8 = 150.9 K
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: P X P = \u22c5 \u00b0 = \u00d7 = 2 0 9 40 36 . mm kg
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: \u0394 = \u22c5 = \u00d7 \u00d7 \u00d7 = T K m b b 0 52 0 1 0 9 18 1000 3 2 . . . . K \u2234 B.P. of solution = 373 + 3.2 = 376.2 K Comprehension IX
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 524 in PDFExtracted text
Solution: Increase in mass of absorber a P \u00b0 and decrease in mass of pure solvent a ( P \u00b0 \u2013 P ). \u2234 P P P X x x x \u00b0 \u2212 \u00b0 = = = + \u2212 0 02 0 24 180 180 100 18 1 . .\n10.38 Chapter 10 HINTS AND EXPLANATIONS \u2234 Mass percent of glucose, x = 1000 21 %
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Solution: AlCl Al Cl 3 3 1 0 8 0 2 0 8 3 0 8 2 4 3 \u2212 = \u00d7 = + \u2212 + . . . . . \u001f \u21c0 \u001f \u21bd \u001f \u001f Total effective mole of solute = 0.2 + 0.8 + 2.4 = 3.4 Now, decrease in mass of solution a P and increase in mass of absorber a P \u00b0. \u2234 P P X m \u00b0 = = + = = \u0394 2 17 17 3 4 5 6 0 18 . . absorber \u2234 Increase in mass of absorber = 0.216 gm
Answer: A
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Solution: Henry\u2019s law
Answer: A
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Answer: D
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Solution: Both have same \u0394 T f
Answer: A
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Solution: KCl will dissociate.
Answer: C
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Answer: D
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Solution: Relative lowering of V.P. is also independent of solvent.
Answer: D
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Solution: Deviation may occur in non-ideal solution.
Answer: A \u2192 Q, B \u2192 P, R, C \u2192 P, R; D \u2192 P, S
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Answer: A \u2192 P, B \u2192 Q, R, C \u2192 R, S; D \u2192 R, S
\nOriginal PDF solution page
Open page 525 in PDFExtracted text
Answer: A \u2192 P, B \u2192 R, C \u2192 P; D \u2192 Q
\nOriginal PDF solution page
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Solution: (A) Some concentrations (B) Osmolarity : NaCl = 0.2 M, Na 2 SO 4 = 0.3 M (C) Osmolarity : NaCl = KCl = 0.2 M (D) Osmolarity : CuSO 4 = 0.2 M, Sucrose = 0.1 M
Answer: A \u2192 P; B \u2192 Q; C \u2192 R; D \u2192 S
\nOriginal PDF solution page
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Solution: Higher the B.P. of solvent, normally higher is its K b value.
Answer: A \u2192 Q; B \u2192 R; C \u2192 S; D \u2192 P
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Solution: (A) 2 = 1 + a (2 \u2013 1) \u21d2 a = 1.00 (B) 2 = 1 + a (3 \u2013 1) \u21d2 a = 0.50 (C) 2 = 1 + a (5 \u2013 1) \u21d2 a = 0.25 (D) 2 = 1 + a (4 \u2013 1) \u21d2 a = 0.33
Answer: A \u2192 Q; B \u2192 P; C \u2192 R; D \u2192 S
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Solution: (A) i = 1 (B) i = 1 + 1(2 \u2013 1) = 2 (C) i = 1 + 1 (3 \u2013 1) = 3 (D) i = 1 + 1(4 \u2013 1) = 4
Answer: A \u2192 P; B \u2192 Q, R, S; C \u2192 T; D \u2192 P, Q, R, S, T
\nOriginal PDF solution page
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Solution: (P) Eff ective conc. = 0.1 \u00d7 3 = 0.3 M = 0.3 m (Q) Eff ective conc. = 0.14 \u00d7 2 = 0.28 M = 0.28 m (R) Eff ective conc. = 0.1 [1+0.9(3 \u2013 1) = 0.28 M = 0.28 m (S) Eff ective conc. = 0.28 M = 0.28 m (T) HA H A M M M ( . ) 0 1 \u2212 + \u2212 + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f K x x x x a = = \u22c5 \u2212 \u21d2 = 0 81 0 1 0 09 . . . \u2234 Eff ective conc. = (0.1 + x ) = 0.19 M = 0.19 m
Answer: 2
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Solution: V gas a n Solvent but independent of pressure. \u2234 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u21d2 = V V V V V V 2 1 2 1 2 2 4 0 5 1 2 gas Solvent ml ml .
Answer: 2
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Solution: m solution = m water + m ethanol or, V \u00d7 0.9344 = 50 \u00d7 1.000 + 50 \u00d7 0.7939 \u2234 V \u2248 96 ml < 100 ml Solution is non-ideal with negative deviation.\n10.39 Liquid Solution HINTS AND EXPLANATIONS
Answer: 2
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Solution: Ideal gas can never be liquefied.
Answer: 1
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Solution: P = X 2 \u22c5 P \u00b0 = 0.8 \u00d7 233.5 =
Answer: 3
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Solution: 8 torr = P exp \u2234 Solution is ideal.
Answer: 2
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Solution: \u03c0 \u03c0 \u03c0 = + + = \u00d7 + \u00d7 + = 1 1 2 2 1 2 2 4 2 4 2 2 3 V V V V V V V V ( ) . . atm
Answer: 5
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Solution: \u03c0 = CRT = \u03c1 g h or 0 2 100 1000 0 0821 300 1 013 1000 0 2463 1 013 10 6 . / / . . . . M \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u2234 M = 2 \u00d7 10 5
Answer: 5
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Solution: \u03c0 = CRT = \u03c1 g h or n 1 0 08 298 1 013 1000 7 45 1 013 10 6 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 . . . . \u2234 n = \u00d7 \u2212 25 10 80 3 \u2234 Millimoles in 320 gm = 25 80 320 20 5 \u00d7 =
Answer: 4
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Solution: V V C C T T 2 1 1 2 1 1 2 2 500 283 105 3 298 5 = = = \u2248 \u03c0 \u03c0 / / / . /
Answer: 2
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 \u21d2 \u2248 T K m n n f f 0 29 1 86 1 04 267 100 1000 4 . . . / /
Answer: 0054
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Solution: \u0394 = \u22c5 T K m f f For KCN solution: 0.80 = K f \u00d7 0.2 \u00d7 2 (1) Hg(CN) mCN Hg(CN mole Final 0 mole (0.2 m mole 2 0 1 0 2 0 1 . . . ) ) + \u2212 \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f m m m + \u2212 2 0 0 1 . mole Final eff ective molality = (0.2 \u2013 0.1 m) + 0.1 + 0.2 = 0.5 \u2013 0.1 m Now, 0.60 = K f \u00d7 (0.5 \u2013 0.1 m) (2) From (1) and (2): m = 2 Four Digit Integer Type
Answer: 0380
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Solution: P X P = \u22c5 \u00b0 2 20 180 18 6 180 18 = + \u00d7 \u00b0 / M P (1) and 20.02 = 11 6 11 M + \u00d7 \u00b0 P (2) \u2234 M = 54
Answer: 0060
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Solution: Mole fraction of solvent is same in both. \u2234 90 18 10 90 18 95 18 5 180 95 18 380 M M + = + \u21d2 = X
Answer: 0795
\nOriginal PDF solution page
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Solution: P P P X \u00b0 \u2212 \u00b0 = = \u2212 = 1 2400 2300 2400 1 24 1 mole solution Urea mole gm Water mole = = \u00d7 = = = 1 24 1 24 60 2 5 23 24 23 2 . 4 4 18 17 25 \u00d7 = \u23a7 \u23a8 \u23aa \u23aa \u23a9 \u23aa \u23aa . gm \u2234 Volume of 1 mole solution = + = 2 5 17 25 1 185 50 3 . . . ml Now, \u03c0 = = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 = CRT 1 24 50 3 1000 0 08 300 60 / / . atm
Answer: 0125
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Solution: \u0394 = \u22c5 T K m f f or, 30 1 86 62 795 1000 795 = \u00d7 \u21d2 = . / / w w gm
Answer: 0058
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Solution: \u0394 = \u22c5 T K m f f or, 6 1 86 50 62 1000 250 = \u00d7 \u21d2 = . / / w w water (final) gm \u2234 Mass of water separated as ice = 375 \u2013 250 = 125 gm
Answer: 0744
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Solution: \u0394 = \u22c5 T K m f f Naphthalene solution: 13 5 38 4 128 185 1000 . . / / = \u00d7 K f (1) Unknown substance solution: 9 0 11 6 185 1000 . . / / = \u00d7 K f M (2) \u2234 M = 58
Answer: 0160
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 = \u00d7 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = \u00b0 T K m f f 1 86 3 6 180 3 6 60 200 1000 0 744 . . . . C \u2234 F. P. of solution = \u2013 0.744\u00b0C\n10.40 Chapter 10 HINTS AND EXPLANATIONS
Answer: 0096
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Solution: \u0394 = \u22c5 T K m f f ( . . ) . / M / 26 84 25 64 8 2 4 10 100 10 1000 3 3 \u2212 = \u00d7 \u00d7 \u00d7 \u21d2 = \u2212 \u2212 M 160
Answer: 0020
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m f f 1 60 4 88 2 122 26 1000 1 1 2 1 . . / / \u03b1 \u2234 a = 0.96 or 96 %
Answer: 0098
\nOriginal PDF solution page
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Solution: \u03c0 = \u21d2 = + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 CRT x 4 92 200 0 05 2 0 08 300 . . . \u2234 x = 20
Answer: 0050
\nOriginal PDF solution page
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Solution: P P X P n n n P n n P \u00b0 \u2212 = \u22c5 \u00b0 = + \u00b0 \u2248 \u22c5 \u00b0 1 1 1 2 1 2 , Urea solution: 0 03 0 1 1000 18 . . = \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00b0 P KCl solution: 0 0594 0 1 1 2 1 1000 18 . . [ ( )] = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00b0 \u03b1 P \u2234 a = 0.98 or 98 %
Answer: 0075
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Solution: Loss in weight of solution a P Loss in weight of water a ( P \u00b0 \u2013 P ) Now, P P P n n \u00b0 \u2212 = \u21d2 = + \u2212 1 2 0 01 0 98 1 25 90 1 3 1 49 18 . . . [ ( )] \u03b1 \u2234 a = 0.50 or 50 %
Answer: 0377
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Solution: \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m f f 7 14 75 2 94 1 1 2 1 . \u03b1 \u2234 a = 0.75 or 75 %
Answer: 0060
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Solution: 100 gm solution (Say) Water = 100 \u2013 (12 + 9.5) = 78.5 gm MgCl2 = 9.5 gm = = 0.1 mole 9.5 9.5 MgSO4 = 12 gm = = 0.1 mole 12 120 Eff ective moles of solute = 0 \u22c5 1 [1 + 0.8(2 \u2013 1)] + 0.1 [1 + 0.6 (3\u2013 1)] = 0.4 Now, \u0394 = \u22c5 = \u00d7 = T K m b b 0 785 0 4 78 5 1000 4 . . . / K \u2234 B. P. of solution = 373 + 4 = 377 K
![\"Nuclear](\"./assets/chapter-packs/nuclear-chemistry/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: Nuclear forces are same in between any two nucleon and it is attractive at 1 fm but repulsive forces are also there between protons
Answer: C
\nOriginal PDF solution page
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Solution: Informative (B.E./nucleon is maximum for Fe)
Answer: B
\nOriginal PDF solution page
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Solution: For lighter nuclei, n p > 1 may make the nucleus unstable
Answer: C
\nOriginal PDF solution page
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Answer: C
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Answer: D
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Solution: Number of n and p , both is even in 30 Zn 64 .
Answer: D
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Solution: r A N \u221d 1 3 / \u21d2 r r 1 2 1 2 = \u00d7 \u21d2 ( ) ( ) / / A 1 1 3 1 3 1 2 56 = \u00d7 \u21d2 A 1 = 7
Answer: B
\nOriginal PDF solution page
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Answer: A
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Solution: For 1 H 1 , n p = = 0 1 0
Answer: B
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Solution: For 1 H 3 , 0 693 32 365 24 730 90 0 693 64 90 . . \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f N w N A A (Some isotopes having n p ratio greater that, 1 H 3 are also know, like 2 He 8 )
Answer: C
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Answer: A
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Solution: Informative Radioactivity
Answer: C
\nOriginal PDF solution page
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Solution: Experimental reason behind considering \u03b1 -particle as He- nucleus.
Answer: A
\nOriginal PDF solution page
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Solution: Experimental fact
Answer: C
\nOriginal PDF solution page
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Solution: Experimental reason behind considering b-emission as nuclear charge.
Answer: D
\nOriginal PDF solution page
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Solution: Isotope formation
Answer: C
\nOriginal PDF solution page
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Solution: Reason of g -emission
Answer: C
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Solution: For an increase in mass, large amount of energy is needed and hence, it is non-spontaneous.
Answer: C
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Solution: b N a N N c \u2212 \u00d7 = \u2212 \u00d7 + \u00d7 = \u03b1 \u03b1 \u03b2 \u03b1 4 2 1 and \u2234 N b N c a b d \u03b1 \u03b2 \u03b1 = \u2212 = \u2212 + \u00d7 \u2212 4 2 4 and ( )
Answer: C
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Solution: n p n p F F \u239b \u239d \u239c \u239e \u23a0 \u239f < \u239b \u239d \u239c \u239e \u23a0 \u239f 18 19 , Hence, F 18 should undergo a -decay on b + - decay on k -capture. Normally, a -decay and k -capture is not found in lighter nuclei.
Answer: C
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Solution: 11 23 10 23 Na Ne \u2192 + F
Answer: A
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Solution: n p n p \u239b \u239d \u239c \u239e \u23a0 \u239f > \u239b \u239d \u239c \u239e \u23a0 \u239f Na Na 24 23 Hence, Na 24 should undergo \u03b2 -decay.
Answer: A
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Solution: c N 14 14 \u2192 \u2212 \u03b2
Answer: D
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Answer: A
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Solution: \u0394 m m m u = \u2212= \u2212 = Au Hg 198 198 197 968 197 966 0 002 . . . \u2234 Q \u2013 value = 0.002 \u00d7 931.5 = 1.8630 MeV But Hg 198 is having energy 1.063 MeV greater than Hg 198 and hence, maximum K.E. of emitted b \u2013particle = 1.863 \u2013 1.063 = 0.8MeV. HINTS AND EXPLANATIONS EXERCISE (JEE ADVANCED)\n14.16 Chapter 14 HINTS AND EXPLANATIONS Rate Law
Answer: C
\nOriginal PDF solution page
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Answer: B
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Solution: r \u221d N \u2032
Answer: D
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Solution: Rate is independent from all external factors.
Answer: D
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Solution: r N = = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u03bb 0 693 28 3 15 10 1 90 6 10 5 24 10 7 23 12 . . . dpspg
Answer: B
\nOriginal PDF solution page
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Solution: r N A 1 0 693 10 10 = \u00d7 \u00d7 . ( ); r N A 2 0 693 5 1 = \u00d7 \u00d7 . ( ) r N A 3 0 693 2 5 = \u00d7 \u00d7 . ( ); r N A 4 0 693 1 2 = \u00d7 \u00d7 . ( )
Answer: C
\nOriginal PDF solution page
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Solution: t \u00bd is independent from amount.
Answer: B
\nOriginal PDF solution page
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Solution: N N n = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 0 1 2
Answer: D
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Solution: w w o n = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 2 \u21d2 3 1 2 12 3 g w o = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 w o = 48gm
Answer: D
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Solution: Moles of He formed = \u00d7 \u00d7 4 5 10 6 10 23 23 . = 0.75 = Moles of decayed \u2234 t t = \u00d7 = 2 2 0 1 2 / hrs
Answer: D
\nOriginal PDF solution page
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Solution: Number of atoms present at time T 1 , N R T 1 1 0 693 = . / Number of atoms present at time T 2 , N R T 2 2 0 693 = . / \u2234 Number of atoms decayed = \u2212 ( ) . R R T 1 2 0 693
Answer: B
\nOriginal PDF solution page
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Solution: Rate should decrease 1 64 1 2 6 = times and hence, t t = \u00d7 = 6 12 1 2 / hrs
Answer: D
\nOriginal PDF solution page
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Solution: P w w w Q w w : : 10 20 40 20 20 1 1 2 0 day 0 day 0 day \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af (As fi nal mass ratio is 1 : 4) Hence, Q is non-radioactive.
Answer: A
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Solution: t \u00bd = 30 min Now, r = \u03bb N \u21d2 28 0 7 30 1200 = \u00d7 \u21d2 = . N N
Answer: C
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Solution: r r o = \u00d7 \u00d7 = = 3 10 3 10 1 8 1 2 8 8 3 \u21d2 t t = \u00d7 = \u00d7 = 3 3 12 26 36 78 1 2 / . . yrs
Answer: C
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Solution: w w o n = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 2 \u21d2 10 1 2 3 6 mg = \u239b \u239d \u239c \u239e \u23a0 \u239f w o / \u21d2 w o = 14 14 . mg
Answer: A
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Solution: r N = = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u03bb 0 693 1 3 10 365 24 3600 75 10 0 35 100 0 012 100 40 6 9 3 . . . . .0 022 10 017 64 23 \u00d7 \u239b \u239d \u239c \u239c \u239c \u239c \u239e \u23a0 \u239f \u239f \u239f \u239f = . dps
Answer: C
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Solution: t t r r t o = \u22c5 \u21d2 > \u22c5 1 2 1 2 2 2 2 5 / / log log log log . \u2234 t 1/2 = 5.25 days
Answer: A
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Solution: Let the sample contains x gm Pu 239 . Now, r N N Pu Pu = + ( ) ( ) \u03bb \u03bb 239 240 or 6 10 0 693 2 4 10 365 24 3600 239 6 022 10 0 693 9 4 23 \u00d7 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + . . . . x 7 7 17 10 365 24 3600 1 240 6 022 10 3 23 . . \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f x \u21d2 x = 0.3112 Hence, mass percent of Pu 239 = \u00d7 = x 1 100 31 12 . %\n14.17 Nuclear Chemistry HINTS AND EXPLANATIONS
Answer: A
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Solution: U Pb 238 206 \u23af \u2192 \u23af Initial a 0 Present a \u2013 x x From question, ( ) . a x x \u2212 \u00d7 \u00d7 = 238 206 1 0 1 \u21d2 x a = 238 2298 Now, age of ore, t t a a x = \u22c5 \u2212 1 2 2 / log log = \u00d7 \u22c5 \u2212 = \u00d7 4 5 10 0 3 1 1 238 2298 7 2 10 9 8 . . log . years
Answer: C
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Solution: Th Pb He 232 208 4 6 \u23af \u2192 \u23af + Initial a mole 0 Present ( a \u2013 x ) mole 6 x mole = \u00d7 = \u00d7 \u2212 \u2212 4 64 10 232 2 10 7 9 . = \u00d7 \u21d2 \u00d7 \u2212 \u2212 6 72 10 22400 5 10 5 10 . \u2234 a = 2.5 \u00d7 10 \u20139 Now, age of sample, t t a a x = \u22c5 \u2212 1 2 2 / log log = \u00d7 \u00d7 \u00d7 \u2212 \u00d7 = \u00d7 \u2212 \u2212 \u2212 1 38 10 0 3 2 5 10 2 5 10 2 5 10 4 6 10 10 9 9 9 9 . . log . . . . years Parallel and Sequential Decay
Answer: B
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Solution: 224 is an integer multiple of 4 and hence, Ra 224 belongs to 4n series, which is thorium series.
Answer: C
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Answer: B
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Answer: D
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Answer: B
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Solution: r r Th Ra = \u21d2 N t N t Th Th Th Ra ( ) ( ) / / 1 2 1 2 = \u21d2 N N Th Ra = 80000 1600
Answer: D
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Solution: (a) \u03bb AC Yr 227 0 693 22 3 15 10 2 1 = = \u00d7 \u2212 \u2212 . . (b) l for the formation of Th 229 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 2 100 3 15 10 6 3 10 2 4 . . Yr (c) l for the formation of Fr 223 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 98 100 3 15 10 3 087 10 2 2 1 . . Yr (d) N N m m Th Fr Th Fe 227 223 227 223 2 98 2 227 98 223 1 49 = \u21d2 = \u00d7 \u00d7 \u2260
Answer: C
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Solution: The net rate of formation of radioisotope, + = \u2212 dn dt R N \u03bb . After very long time, steady state will be achieved, at which + = dn dt
Answer: A
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Solution: Hence, N R = \u03bb .
Answer: C
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Solution: Pb Bi hr hr 212 8 212 1 1 2 1 2 t t / / = = \u23af \u2192 \u23af\u23af\u23af \u23af \u2192 \u23af\u23af\u23af Time for maximum nuclei and hence, maximum activity of Bi212, t max ln = \u2212 \u22c5 1 2 1 2 1 \u03bb \u03bb \u03bb \u03bb = = \u22c5 = = 1 2 1 2 8 1 1 1 8 3 429 205 7 ln ln ln . . min hr
Answer: B
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Solution: \u03bb \u03bb \u03bb \u03b1 \u03b2 overall = + \u21d2 = + 0 693 0 693 20 0 693 60 1 2 . . . / t \u2234 t 1/2 = 15 min \u2234 For 87.5 % decay, t t = \u00d7 = 3 4 5 1 2 / min
Answer: D
\nOriginal PDF solution page
Open page 690 in PDFExtracted text
Solution: Average energy released = \u00d7 + \u00d7 + = 0 05 40 0 15 80 0 05 0 15 70 . . . . MeV\n14.18 Chapter 14 HINTS AND EXPLANATIONS Nuclear Reactions
Answer: D
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 92 235 0 1 54 139 38 94 0 1 3 U n Xe Sr n + \u23af \u2192 \u23af + +
Answer: A
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 25 55 0 1 25 56 Mn n Mn + \u23af \u2192 \u23af + \u03b3
Answer: C
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 4 9 1 1 5 10 Be H B (proton) + \u23af \u2192 \u23af + \u03b3
Answer: C
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 13 27 2 4 15 30 0 1 Al He P n particle + \u23af \u2192 \u23af + \u2212 ( ) \u03b1
Answer: A
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Answer: D
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Answer: B
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Answer: A, C, D
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Answer: A, D
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: Q -value is distributed between \u03b2 -particle and anti- neutrino.
Answer: C
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 92 235 90 23 88 227 89 227 93 235 U Th Ra Ac Np \u2212 \u2212 \u2212 \u2212 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u03b1 \u03b1 \u03b2 \u03b2 \u03b1 9 91 231 Pa \u2212 89 AC 235 is not possible.
Answer: A, B, C
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: Activity is independent from all external factors.
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: Half-life of a radio isotope is its characteristic property, independent from all factors.
Answer: A
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: Actinum series: 92 235 82 207 U Pb \u2192
Answer: A
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Answer: B, C, D
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 96 242 2 4 97 293 0 1 2 Cm He Bk Incorrect + \u23af \u2192 \u23af + n ( ) 5 10 2 4 7 13 0 1 7 19 0 1 6 14 1 1 1 B He N n Correct N n C H (Correct) + \u23af \u2192 \u23af + + \u23af \u2192 \u23af + ( ) 9 9 28 1 2 15 29 0 1 Si H P n (Correct) + \u23af \u2192 \u23af +
Answer: C
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: Neutron is projectile and proton is emitted particles.
Answer: D
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 13 27 2 4 15 30 0 1 Al He P n + \u23af \u2192 \u23af + 6 96 C He N P Si e Au He BK 12 1 1 7 13 15 30 14 30 +1 0 241 2 4 97 24 + \u23af \u2192 \u23af + \u23af \u2192 \u23af + + \u23af \u2192 \u23af \u03b3 4 4 1 1 H +
Answer: A, B
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 4 9 4 8 0 1 Be Be n + \u23af \u2192 \u23af + \u03b3 4 9 1 1 4 8 1 2 Be H Be H + \u23af \u2192 \u23af +
Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: \u0394 m u = \u00d7 + \u00d7 \u2212 = ( . . ) . 8 1 0072 8 1 0086 16 0 1264 \u2234 B.E. per nucleon = \u00d7 = 0 1264 931 5 16 7 36 . . . MeV
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 8 16 2 4 4 O He \u23af \u2192 \u23af \u0394 m u = \u2212 \u00d7 = \u2212 15 9944 4 4 0026 0 016 . . . \u2234 Energy required in separation = \u00d7 = 0.016 MeV 931 5 14 904 . .
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 691 in PDFExtracted text
Solution: 10 20 6 12 2 4 2 Ne C He \u23af \u2192 \u23af + Energy required = \u00d7 \u2212 \u00d7 + \u00d7 \u00d7 (20 8 03 12 7 68 2 4 7 07 . ) ( . . ) = 11.88 MeV\n14.19 Nuclear Chemistry HINTS AND EXPLANATIONS Comprehension II
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: SC 50 50 \u23af \u2192 \u23af + + \u03c4 \u03b2 \u03bd i Q \u2212 = \u2212 \u00d7 = value (49.9516 49.94479 MeV ) . . 931 5 6 34 \u2234 K.E. of MeV \u03bd = \u2212 = 6 34 0 80 5 54 . . .
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: \u03bb = = \u00d7 \u00d7 \u00d7 \u2212 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 LC E \u0394 6 626 10 3 10 4 795 4 611 10 1 6 10 6 75 1 39 8 6 19 . ( . . ) . . 0 0 12 \u2212
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: Th Ra Ra 228 224 224 \u2212 \u2212 \u23af \u2192 \u23af \u23af \u2192 \u23af \u03b1 \u03b3 Q-value = 228 028726 224 020196 4 0026 4 0026 931 6 . . . . . ( ) ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 + + \u00d7 \u2212 2 217 10 3 5 307 \u00d7 \u2212 = . MeV \u2234 \u00d7 = K.E. of -particle = MeV \u03b1 224 228 5 307 5 214 . . Comprehension III
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: Overall rate is the rate of slowest step and hence, T required = 270 days
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: At transient equilibrium, N N A B B A A = \u2212 \u22c5 \u03bb \u03bb \u03bb or, N N N N A B Th Ra Ra Th Th = = \u2212 = \u2212 \u00d7 \u00d7 = \u03bb \u03bb \u03bb ln . ln . ln . 2 3 64 2 1 913 365 2 1 913 365 190 0 8 1 .
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: At secular equilibrium, N N B A B A = \u22c5 \u03bb \u03bb or, N N N N A B Ra Rn Rn Ra = = = \u00d7 \u00d7 = \u03bb \u03bb ln ln . . 2 55 2 3 65 24 3600 5733 8 Comprehension IV
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: U Pb 238 206 \u23af \u2192 \u23af Initial a mole 0 Present ( a \u2013 x ) mole x mole = 59 5 238 . = \u00d7 12 875 206 80 100 . \u2234 a = 0.30 Now, t a a x = \u22c5 \u2212 = \u00d7 \u22c5 \u2212 1 1 1 52 10 0 3 0 25 10 \u03bb ln . ln . . = 1.33 \u00d7 10 9 Yrs
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: K Ar 40 40 \u23af \u2192 \u23af Initial a mole 0 Present ( a \u2013 x ) mole x mole = 1 = 10.3 \u2234 a = 11.3 Now, t t a a x = \u22c5 \u2212 = \u00d7 \u22c5 1 2 9 2 1 25 10 0 3 11 3 1 / log log . . log . = 4.375 \u00d7 10 9 years
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 692 in PDFExtracted text
Solution: U Pb 238 206 \u23af \u2192 \u23af K Ar 40 40 \u23af \u2192 \u23af Initial mole 0 b mole 0 Present ( a \u2013 x ) mole x mole ( b \u2013 y ) mole y mole = \u00d7 \u2212 0 86 10 238 3 . = \u00d7 \u2212 0 15 10 206 3 . = \u00d7 \u2212 10 10 40 3 = \u00d7 \u2212 1 6 10 40 3 . t t a a x t b b y U K = \u22c5 \u2212 = \u22c5 \u2212 ( ) log log ( ) log log / / 1 2 1 2 238 40 2 2 \u2234 w = 1.7 mg\n14.20 Chapter 14 HINTS AND EXPLANATIONS Comprehension V
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Given in paragraph
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: For radioactive tracing, time should be comparable to t 1/2 .
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: T c c T c c 1 1 2 2 1 1 = \u22c5 = \u22c5 \u03bb \u03bb ln ln and As c c T T T T c c 1 2 1 2 1 2 1 2 1 > > \u2212 = \u22c5 , ln and \u03bb Comprehension VI
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Let the volume of blood be V ml. t T N N N av 1 0 0 1 2 = = \u22c5 \u2212 ln \u21d2 5 = \u22c5 15 2 1260 15 60 log log / V \u2234 V = 4000
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: r 0 1260 60 4000 18 9 = \u00d7 = . dpm per ml Now, r r r r 0 5 5 10 = \u21d2 18 9 15 15 10 . = r \u21d2 r 10 = 11.9 dpm per ml Comprehension VII
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Isotopes are B and E, C and F, D and G.
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Mass number of H = 230 \u2013 4 \u00d7 4 = 214
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Z A \u2013 4 \u00d7 2 + 3 \u00d7 1 = 88 \u21d2 Z A = 93 Comprehension VIII
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Net rate of formation + = \u2212 dn dt N \u03b1 \u03bb or, dN N dt N N t \u03b1 \u03bb \u2212 = \u222b \u222b 0 0 \u21d2 N N e t = \u2212 \u2212 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 1 0 \u03bb \u03b1 \u03b1 \u03bb \u03bb ( ).
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: t t N = = = 1 2 0 2 2 / ln \u03bb \u03b1 \u03bb and \u2234 N = 1.5 N 0
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: t N N \u2192 \u221e = = , then \u03b1 \u03bb 2 0 Comprehension IX
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: \u03bb \u03bb \u03bb = + 1 2 \u21d2 ln ln ln / 2 2 24 2 8 1 2 t = + \u21d2 t 1/2 = 6 hours.
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Activity of excreted material in 48 hours N N T av 1 0 0 116 16 = = . and sec. But as T c is simultaneously decaying with t 1/2 = 8 hrs. Final activity after 48 hrs = = 24 2 0 375 6 . . \u03bc ci
Answer: C
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: n p ratio does not increases continuously.
Answer: C
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Binding energy increases but the binding energy per nucleon first increases and then decreases.
Answer: A
\nOriginal PDF solution page
Open page 693 in PDFExtracted text
Solution: Stable\n14.21 Nuclear Chemistry HINTS AND EXPLANATIONS
Answer: A
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Answer: D
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Solution: All heavy nuclei should not produce 82 Pb 206 .
Answer: A
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Answer: C
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Solution: \u03b2 -decay occurs to decrease n p ratio.
Answer: D
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Solution: Same mass of U 238 and U 238 F 6 have diff erent numbers of U 238 nuclei.
Answer: A
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Solution: t T av 1 2 0 693 1 / . / / = \u03bb \u03bb = 0.693 = Same for all
Answer: C
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Solution: Mesons have mass 200 to 300 times mass of electrons.
Answer: A
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Answer: B
\nOriginal PDF solution page
Open page 694 in PDFExtracted text
Solution: 13 Ae 30 have high n p ratio than its stable nucleus 13 Ae 27 .
Answer: A
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Answer: D
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Answer: A \u2192 R; B \u2192 Q; C \u2192 P; D \u2192 S
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Answer: A \u2192 P; B \u2192 Q; C \u2192 R; D \u2192 P; E \u2192 P
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Answer: A \u2192 Q, S; B \u2192 P, T; C \u2192 R, T; D \u2192 T; E \u2192 T
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Answer: A \u2192 S; B \u2192 Q; C \u2192 R; D \u2192 P
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Solution: 53 I 127 is stable and hence, I 333 is beta emitter and I 121 is positron emitter.
Answer: A \u2192 S; B \u2192 R; C \u2192 Q; D \u2192 P
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Solution: (a) 92 U 235 82 Pb 207 N N \u03b1 \u03b2 = \u2212 = = \u2212 \u2212 \u00d7 = 235 207 4 7 82 92 2 7 4 ( ) (b) 92 U 238 82 Pb 206 N N \u03b1 \u03b2 = \u2212 = = \u2212 \u2212 \u00d7 = 238 206 4 8 82 92 2 8 6 ( ) (c) 94 Pu 241 83 Bi 209 N N \u03b1 \u03b2 = \u2212 = = \u2212 \u2212 \u00d7 = 241 209 4 8 83 94 2 8 5 ( ) (d) 90 Th 232 82 Pb 208 N N \u03b1 \u03b2 = \u2212 = = \u2212 \u2212 \u00d7 = 232 208 4 6 82 90 2 6 4 ( )
Answer: 2
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Solution: Number of half-lifes 28 1 2 81 10 . . = \u2234 Mass of Sr 90 remained = \u00d7 = \u00d7 \u2212 2 048 10 2 2 10 10 6 . gm gm
Answer: 3
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Solution: r N = = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u03bb 0 7 14 24 3600 86 4 10 164 0 164 100 6 10 3 10 3 23 . . . 1 11 dps
Answer: 9
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Solution: Z m Z m A B \u23af \u2192 \u23af + \u2212 \u2212 6 12 2 4 3 He t = 0 1 mole 0 t = 20 days 1 3 4 \u2212 3 3 4 \u00d7 mole = 1 4 mole \u2234 V = \u00d7 9 4 22.4 = 9 \u00d7 5.6 L at 0\u00b0C and 1 atom
Answer: 1
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Solution: 1 2 0 335 2 mV eV = .\n14.22 Chapter 14 HINTS AND EXPLANATIONS \u2234 V = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 2 0 335 1 6 10 1 675 10 8000 19 27 . . . m/s Hence, time for travelling 80 km, t d v = = \u00d7 = 80 10 8000 10 3 sec Now, t t N N = \u22c5 1 2 0 2 / ln ln or, 10 700 2 100 100 = \u22c5 \u2212 ln x \u21d2 x = 0.99 \u2248 1
Answer: 6
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Solution: t t N N t N N U = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 2 0 1 2 0 2 2 238 235 / / log log log log U or, 4 5 10 2 140 7 2 10 2 9 0 8 0 . log log . log log \u00d7 \u22c5 = \u00d7 N x N x \u2234 log . N x 0 2 5 = \u2234 Age of earth, t N x = \u00d7 \u22c5 = \u00d7 7 2 10 2 6 10 8 0 9 . log log years Four-digit Integer Type
Answer: 0535
\nOriginal PDF solution page
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Solution: t t r r = \u22c5 1 2 0 2 / log log or, 6 93 6 93 2 5 10 0 15 . . log log = \u22c5 \u00d7 r \u21d2 r 0 = 5.35 \u00d7 10 15 dpm Now, r N 0 0 = \u03bb . or, 5.35 \u00d7 = \u00d7 \u00d7 \u00d7 \u00d7 10 0 693 69 3 60 6 10 15 23 . . ( ) n \u2234 n = 5.35 \u00d7 20 \u20135
Answer: 0060
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Solution: Initial number of H 3 atoms = 0 93 10 18 6 10 2 8 10 3 23 18 . \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 = 4.8 \u00d7 10 2 Number of half lives = = 36 9 12 3 3 . . \u2234 Final number of H 3 atoms
Answer: 0016
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Solution: t T N N N av 1 0 0 1 2 = = \u22c5 \u2212 ln and 3 6 8 3 1 0 0 1 t T N N N av = = \u22c5 \u2212 ln \u2234 N N T av 1 0 0 116 16 = = . and sec
Answer: 0050
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Solution: Let the mass of water present in body = w gm. Now, 9 \u00d7 10 9 = 2.25 \u00d7 10 5 \u00d7 w \u21d2 w = 4 \u00d7 10 4 gm = 40 kg \u2234 Mass per cent of water in body = \u00d7 = 40 80 100 50%
Answer: 0040
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Solution: \u03bb \u03bb \u03b2 \u2212 = \u00d7 32 100 overall \u2234 t t 1 2 1 2 100 32 100 32 12 8 40 / / . ( ) = \u00d7 ( ) = \u00d7 = \u2212 \u03b2 overall hr
Answer: 0449
\nOriginal PDF solution page
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Solution: \u03bb = + 1 1620 1 405 \u21d2 \u03bb = \u2212 1 324 1 Yr \u2234 t t required = \u00d7 = \u00d7 \u00d7 2 2 0 693 324 1 2 / . = 449.064 years
Answer: 0167
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Solution: Sr Y 90 90 \u23af \u2192 \u23af \u23af \u2192 \u23af other format For radioactive equilibrium, ( .N) ( .N) Sr Y 90 90 \u03bb \u03bb = or, 0 693 32 365 24 730 90 0 693 64 90 . . \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f N w N A A \u2234 w = 0.1667 gm
Answer: 0216
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Solution: Energy released = 2 \u00d7 120 \u00d7 8.1 \u2013 240 \u00d7 7.2 = 216 MeV\n14.23 Nuclear Chemistry HINTS AND EXPLANATIONS
Answer: 0960
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Solution: \u0394 m = 2 \u00d7 2.0021 \u2013 4.0026 = 0.0016 amu Now, let n moles of H 2 be required. n 2 6 10 0 0016 1 5 10 25 100 200 10 3600 24 23 10 6 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 . . \u2234 n = 960
Answer: 0002
\nOriginal PDF solution page
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Solution: After 2 8 2 2 = half-life, detectable activity = 100 2 % But actual detected activity is 10 % . Hence, mass of iodine migrated in thyroid gland = \u00d7 = 10 2 10 100 2 2 mg
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\nOriginal PDF solution page
Open page 476 in PDFExtracted text
Solution: a b c \u2260 \u2260 = = = \u00b0 , \u03b1 \u03b2 \u03b3 90 \u21d2 Orthorhombic
Answer: B
\nOriginal PDF solution page
Open page 476 in PDFExtracted text
Solution: Volumeof metal taken cm 3 = = = m d 100 6 25 16 . Volumeof each unit cell cm cm 3 = \u00d7 ( ) = \u00d7 \u2212 \u2212 4 10 64 10 8 3 24 \u2234 Number of unit cells = \u00d7 = \u00d7 \u2212 16 64 10 2 5 10 24 23 .
Answer: A
\nOriginal PDF solution page
Open page 476 in PDFExtracted text
Solution: d Z M N V = \u22c5 \u22c5 A \u21d2 45 16 27 6 10 4 10 23 8 3 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) \u2212 Z \u21d2 Z = 4 Hence, Al crystal is FCC. For FCC: 2 4 a r = \u21d2 r = \u00d7 = 2 4 0 4 1 414 . . \u00c5
Answer: B
\nOriginal PDF solution page
Open page 476 in PDFExtracted text
Solution: d Z M N V FCC A 3 gm/cm = \u22c5 \u22c5 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 4 56 6 10 4 125 10 2 8 45 23 10 3 . d Z M N V BCC A 3 gm/cm = \u22c5 \u22c5 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 2 56 6 10 4 50 2 10 3 42 87 23 10 3 . As the density of iron is increased, there is contraction.\n9.30 Chapter 9 HINTS AND EXPLANATIONS
Answer: C
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: Packing fraction = \u00d7 ( ) \u00d7 \u00d7 \u00d7 ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 6 10 4 3 144 10 108 10 6 0 7 23 10 3 \u03c0 . . 3 36 Hence, the crystal should be FCC.
Answer: B
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: d Z M N V = \u22c5 \u22c5 A \u21d2 10 5 4 198 5 10 8 3 . = \u00d7 \u00d7 \u00d7 ( ) \u2212 N A \u21d2 N A = \u00d7 6 034 10 23 .
Answer: D
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: Void space per unit cell = 0.26 \u00d7 V unit cell = \u00d7 ( ) = 0 26 4 16 64 3 3 . . \u00c5 \u00c5
Answer: C
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: d Z M N V = \u22c5 \u22c5 A \u21d2 12 5 4 6 10 4 100 2 10 2 23 10 3 . gm cm gm cm 3 3 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 M \u2234 M = 120 \u21d2 Metal is Sn.
Answer: A
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: P.F. / / particle unit cell particle unit cell particle par d V V m V V m = = t ticle or, 0 1 4 3 1 0 10 6 10 8 3 23 . . / = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u2212 Z Z M \u03c0 \u21d2 M = 8 \u03c0
Answer: B
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: Fraction of edge covered by atoms = = = 2 2 4 2 0 707 r a r r / . Hence, fraction of edge not covered = 0.293
Answer: C
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: d Z M N V = \u22c5 \u22c5 A \u21d2 8 3 4 6 10 5 10 23 8 3 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) \u2212 M \u21d2 M = 50 gm/mol At 0\u00b0C, the substance will exist as gas and its density = = 50 22 4 2 23 gm L g/L . .
Answer: B
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: M HCl MCl 1 2 H 2 + \u2192 + 1 1 2 mole mole = A gm = \u00d7 \u00b0 1 2 22 7 0 1 . L at C and bar \u2234 (7.68 \u00d7 4.5) gm 11 35 7 68 4 5 4 54 . . . . A \u00d7 \u00d7 ( ) = \u2234 A = 86.4 Now, d Z M N V = \u22c5 \u22c5 A \u21d2 4 5 86 4 6 10 400 10 23 10 3 . . = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) \u2212 Z \u2234 Z = 2 \u21d2 Unit cell is BCC.
Answer: C
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: Percentage of occupied space in water = \u00d7 = x x 0 99 0 96 33 32 . . \u2234 Percentage of empty space in water is 100 33 32 \u2212 x .
Answer: B
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: (Volume of crystal containing one mole metal) \u00d7 70 100 = \u00d7 ( ) \u00d7 \u00d7 \u00d7 ( ) \u2212 6 10 4 3 0 2 10 23 7 3 \u03c0 . cm \u2234 V crystal cm = 64 7 3 \u03c0 \u2234 Density gm/cm = \u239b \u239d \u239c \u239e \u23a0 \u239f = 32 64 7 3 5 3 \u03c0 \u03c0 .
Answer: D
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: There is no octahedral voids in BCC. However, all the face centres are distorted octahedral voids, which are not considered because they are not regular voids.
Answer: C
\nOriginal PDF solution page
Open page 477 in PDFExtracted text
Solution: None of the tetrahedral as well as octahedral voids will be in contact with other tetrahedral and octahedral voids, respectively.\n9.31 Solid State HINTS AND EXPLANATIONS
Answer: C
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: Packing is FCC for which 2 4 a r = . \u2234 r = \u00d7 = 2 10 4 2 5 2 \u00c5 \u00c5 . Now, density of metal atom = m V atom atom = \u00d7 ( ) \u00d7 \u00d7 \u00d7 ( ) = \u2212 60 22 6 022 10 4 3 2 5 2 10 0 54 23 8 3 . . . . \u03c0 gm/cm 3
Answer: B
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: P.F. particle unit cell = = \u00d7 \u00d7 \u00d7 ( ) \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa \u00d7 = V V r r r 3 4 3 6 3 4 2 2 3 3 3 2 \u03c0 \u03c0
Answer: B
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: 1 2 4 3 3rd layer 2nd layer 1st layer 1 4 3 2
Answer: C
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: A B A 0.V 0.V 0 h 3 h 4 h 2 h 4
Answer: D
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: Number of NaCl formula units in 1 gm = \u00d7 \u00d7 ( ) 1 58 5 6 10 23 . Each unit cell contains 4 NaCl formula units and hence, volume of crystal = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 6 10 58 5 4 4 7 10 0 12 23 23 . . . ml
Answer: C
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: d Z M N V = \u22c5 \u22c5 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) = \u2212 A 3 gm/cm 4 58 5 6 10 600 10 1 8 23 10 3 . .
Answer: A
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: The structure is simple cubic for both metals.
Answer: A
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: \u2018B\u2019 should occupy all the tetrahedral voids and hence, its C.N. =
Answer: B
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: 25. P.E. A ( ) = \u00d7 + \u00d7 ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f = 4 4 3 8 4 3 0 225 4 2 0 76 3 3 3 \u03c0 \u03c0 r r r . . P.E. B ( ) = \u00d7 + \u00d7 ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f = 4 4 3 4 4 3 0 414 4 2 0 79 3 3 3 \u03c0 \u03c0 r r r . . P.E. C ( ) = \u00d7 + \u00d7 ( ) ( ) = 1 4 3 1 4 3 0 732 2 0 72 3 3 3 \u03c0 \u03c0 r r r . .
Answer: B
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: Z Z x Z O X Y 2 2+ 3+ \u2212 = = \u00d7 = \u00d7 = 4 8 100 4 50 100 2 ; ; For neutrality of crystal, 4 2 8 100 2 2 3 0 \u00d7 \u2212 ( ) + \u00d7 + ( ) + \u00d7 + ( ) = x \u21d2 x = 12.5
Answer: D
\nOriginal PDF solution page
Open page 478 in PDFExtracted text
Solution: r r x Rb Cl pm + \u2212 + = = 328 5 . r r y r r z r r w K Cl Na Br K Br pm pm pm + \u2212 + \u2212 + \u2212 + = = + = = + = = 313 9 298 1 329 3 . . . \u2234 r r x w y Rb Br pm + \u2212 + = + \u2212 = 343 9 .\n9.32 Chapter 9 HINTS AND EXPLANATIONS
Answer: A
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: r r + \u2212 = 0 414 . \u21d2 r \u2212 = = 200 0 414 483 1 . . pm
Answer: B
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: Z Z Z A B C = = \u00d7 = = \u00d7 = 4 2 1 2 1 4 1 4 1 ; ; \u2234 Formula = A 4 BC
Answer: A
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: Z Z Z O Metal I M Metal II N = = \u00d7 = = \u00d7 = ( ) ( ) 4 1 8 8 1 1 2 4 2 ; \u2234 Formula = MN 2 O 4 For neutrality, M Zn and N Al 3+ = = + 2
Answer: A
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: Z M 3+ If M are at corners = \u00d7 = ( ) + 8 1 8 1 3 Z X If F are at face centres \u2212 = \u00d7 = ( ) \u2212 6 1 2 3
Answer: A
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: Octahedral voids in FCC = 4, but only one is occupied by \u2018 x \u2019 and one by \u2018 y \u2019.
Answer: D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: (II) r r + \u2212 = = \u2212 ( ) \u21d2 = 20 95 0 21 0 155 0 225 3 . . . C.N.
Answer: B
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: 0.V. T.V. T.V. Fraction of body diagonal covered = + \u00d7 + \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 2 2 0 414 4 0 225 3 4 2 0 76 r r r r . . . \u2234 Fraction, not covered = 1 \u2013 0.76 = 0.24
Answer: A
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: For electrical neutrality, A = bivalent and B = trivalent. Now, one octahedral void is occupied by \u2018A\u2019 and one by \u2018B\u2019.
Answer: A
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: Frenkel defect does not change the density.
Answer: B
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: ln f f H R T T 2 1 1 2 1 1 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u0394 or, ln / / 1 2 10 1 10 1 1100 1 1200 9 10 \u00d7 ( ) ( ) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u0394 H R \u21d2 \u0394 H = 176 8 . KJ/mol
Answer: D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Answer: D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Answer: A, C
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Answer: C, D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Answer: C, D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Answer: A, D
\nOriginal PDF solution page
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Solution: A B A 0.V. T.V. T.V. T.V. T.V. 0.V 0 h 7 h 8 6 h 8 5 h 8 4 h 8 3 h 8 2 h 8 h 8
Answer: A, B, D
\nOriginal PDF solution page
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Solution: Shortest distance between two T.V. is a 2 .
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: Both have same packing efficiency and C.N.
Answer: A, B, D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: There are the voids per sphere in hexagonal close packing.
Answer: C, D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: For octahedral void, the orientation of both tetrahedral voids should be opposite to each other.
Answer: A, C, D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: a r r = + ( ) + \u2212 2 Na Cl
Answer: B
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Answer: B, C, D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 479 in PDFExtracted text
Solution: Informative\n9.33 Solid State HINTS AND EXPLANATIONS
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Solution: (a) P P c K b H = + + ( ) 7 1 2 log \u21d2 5 0 7 1 2 4 4 . . log = \u2212 + ( ) c \u2234 c M M = = \u21d2 = 0 4 80 2 100 . / (b) h K K c h b = \u00d7 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 10 4 10 0 4 2 5 10 14 5 5 . . (c) 3 2 2 160 186 4 3 400 a r r a x y = + ( ) \u21d2 = \u00d7 + ( ) = + \u2212 . pm (d) d M N V = \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 ( ) = \u2212 Z gm/cm A 3 1 100 6 10 400 10 2 6 23 10 3 .
Answer: A, B
\nOriginal PDF solution page
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Solution: (a) Z Z Z C A B 3 2 4 4 1 2 8 4 \u2212 + + = = = \u00d7 = , , Crystal is electrically neutral and hence, it is possible. (b) Z Z Z B A C 2 3 6 6 1 2 12 6 + + \u2212 = = = \u00d7 = , , Crystal is electrically neutral and hence, it is possible. (c) Z Z Z A B C + + \u2212 = \u00d7 = = \u00d7 = = 4 1 8 1 2 4 1 8 1 2 1 2 3 , , Crystal is negatively changed and hence, it is not possible. (d) Z Z Z B C A 2 3 4 8 4 + \u2212 + = = = , , Crystal is negatively changed and hence, it is not possible.
Answer: A, B, D
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Answer: B, C
\nOriginal PDF solution page
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Solution: CaF Al O 2 2 3 8 4 6 4 : , : ( ) ( )
Answer: B, C, D
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Answer: B, C
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Answer: C
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Solution: Metal defi cient defect can occur with extra anion present in the interstitial voids, but it is very rare.
Answer: B, C
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Answer: A, B, C
\nOriginal PDF solution page
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Solution: \u0394 H = + ve and hence, the surroundings must lose heat.
Answer: A, C
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Solution: O CCP 2 \u2212 = Fe O.V. 2 1 4 1 + = \u00d7 = Fe in T.V. and 1 in O.V. 3 1 + =
Answer: B, C
\nOriginal PDF solution page
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Answer: C, D
\nOriginal PDF solution page
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Answer: C, D
\nOriginal PDF solution page
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Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Solution: 3 4 2 3 5 0 2 4 33 a r r = \u21d2 = \u00d7 = . . \u00c5
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Solution: Next nearest neighbours are at \u2018 a \u2019 distance.
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Comprehension 1 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Solution: Number of next nearest neighbours = 6
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 480 in PDFExtracted text
Solution: d Z M N V = \u22c5 \u22c5 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) = \u2212 A 3 gm/cm 2 39 6 10 5 10 1 04 23 8 3 .
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: Fractional space occupied by atoms = = V V V V atoms liquid atoms/mol liquid/mol = \u00d7 ( ) \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f ( ) = \u2212 6 10 4 3 3 5 10 4 39 0 9 0 5883 23 8 3 \u03c0 / . . \u2234 Percentage of empty space = (1 \u2013 0.5883) \u00d7 100 = 41.17 %\n9.34 Chapter 9 HINTS AND EXPLANATIONS Comprehension II
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 481 in PDFExtracted text
Solution: 2 4 2 0 5 2 4 0 125 a r r = \u21d2 = \u00d7 = . . nm Now, size of octahedral void = 0.414 \u00d7 0.0125 = 0.052 nm
Comprehension 2 - subquestion 4
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Size of tetrahedral void = 0.225 \u00d7 0.0125 = 0.028 nm Comprehension III
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 481 in PDFExtracted text
Solution: Volume Mass Density cm 3 = = \u00d7 \u00d7 ( ) \u00d7 = \u00d7 \u2212 6 24 6 10 1 92 1 25 10 23 22 . .
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Volume occupied by particles = \u00d7 \u03c0 3 2 V unit all or, 6 4 3 3 2 1 25 10 1 5625 10 3 22 8 \u00d7 = \u00d7 \u00d7 \u21d2 = \u00d7 \u2212 \u2212 \u03c0 \u03c0 r r . . cm
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Height of unit cell = \u22c5 = 4 2 3 5 r \u00c5
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Number of nearest neighbours = 12 Comprehension IV
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Number of T.V. per particle = 2 Number of O.V. per particle = 1
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: T.V. are smaller than O.V. Comprehension V
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: a a r r r r r r r r r r KCl NaCl K Cl Na Cl K Na Cl Na Cl N = + ( ) + ( ) = + + + \u2212 + \u2212 + + \u2212 + \u2212 2 2 1 a a + = + + = 1 0 7 1 0 5 1 1 0 5 1 143 . . . .
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: d d a a NaCl KCl NaCl KCl = = \u00d7 ( ) = 58 5 74 5 58 5 74 5 1 143 1 172 3 3 3 . . . . . . Comprehension VI
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Centre is octahedral void.
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Number of O.V. = 4, but only one is occupied.\n9.35 Solid State HINTS AND EXPLANATIONS Comprehension VII
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: C.N. of Zn 2+ = 4
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: C.N. of S 2 \u2212 = 4
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Zn 2+ is tetrahedrally linked with 4 S 2 \u2212 ions.
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: r r + \u2212 > 0 225 . Comprehension VIII
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Z Mn = \u00d7 = 8 1 8 1 Z F = \u00d7 = 12 1 4 3 \u2234 Formula = MnF 3
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: a r r = + ( ) = + ( ) = + \u2212 2 2 0 65 1 35 4 00 3 Mn F . . . \u00c5
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: d = \u00d7 + \u00d7 ( ) \u00d7 ( ) \u00d7 \u00d7 ( ) = \u2212 1 55 3 19 6 10 4 10 2 92 23 8 3 . gm/cm 3 Comprehension IX
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: V m d formula unit 3 cm = = + ( ) \u00d7 \u00d7 = \u00d7 \u2212 132 5 35 5 6 10 3 5 8 10 23 23 . . .
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: d Z M N V A = . . \u21d2 3 5 1 168 6 10 23 3 . = \u00d7 \u00d7 ( ) \u00d7 a \u21d2 a = \u00d7 \u2212 4 3 10 8 . cm Hence, nearest Cs \u2013 Cs distance = a = 4.3 \u00c5
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Nearest Cs Cl distance \u2212 = = 3 2 3 72 a . \u00c5 Comprehension X
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: N N e e O E RT = = = \u00d7 \u2212 \u2212 \u00d7 \u00d7 \u00d7 \u2212 2 46 10 2 2 1000 5 3 1 0 10 .
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: In NaCl, all O.V. are occupied. Hence, the available voids are only tetrahedral. \u2234 N i = 2 \u00d7 N o Now, N N N e o o E RT = \u00d7 \u22c5 \u2212 2 2 / \u2234 N N e e o E RT = \u00d7 = \u00d7 = \u00d7 \u2212 \u2212 \u00d7 \u00d7 \u00d7 \u2212 2 2 1 41 10 2 73 6 1000 2 2 1000 8 / . . Comprehension XI
Comprehension 10 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: d Z M N V = \u22c5 \u22c5 A = \u00d7 \u00d7 ( ) \u00d7 \u22c5 \u00d7 ( ) = = \u2212 4 6 023 6 023 10 2 10 1 200 5 23 1 3 7 3 . . / Y Y gm/cm kg/m 3 3
Comprehension 11 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: d observed >> d theoretical Such large diff erence is possible due to impurity defect.\n9.36 Chapter 9 HINTS AND EXPLANATIONS Comprehension XII
Comprehension 11 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: For diamond crystal, 3 8 a r = and Z = 8 Now, d Z M N V = \u22c5 \u22c5 A \u2234 3 6 8 12 6 10 8 3 23 3 . = \u00d7 \u00d7 ( ) \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f r \u2234 r = 0.76 \u00d7 20 \u20138 cm
Comprehension 11 - subquestion 3
",
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+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: For SiC, 3 4 a r r c si = + ( ) and Z = 4 Now, d Z M N V r r c si = \u22c5 \u22c5 \u21d2 = \u00d7 \u00d7 ( ) \u00d7 + ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f A 3 2 4 40 6 10 4 3 23 3 . d Z M N V r r c si = \u22c5 \u22c5 \u21d2 = \u00d7 \u00d7 ( ) \u00d7 + ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f A 3 2 4 40 6 10 4 3 23 3 . \u2234 r si = \u00d7 \u2212 1 12 10 8 . cm
Comprehension 12 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Interchange between C and si atoms will neither change \u2018 Z \u2019 nor \u2018 a \u2019.
Comprehension 12 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: For the same volume, m m dia sic 3 6 3 2 . . = or, n n n n c sic sic c \u00d7 = \u00d7 \u21d2 = 12 3 6 40 3 2 1 3 75 . . .
Comprehension 12 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: Packing efficiency of SiC is greater due to unequal size of particles.
Answer: C
\nOriginal PDF solution page
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Solution: Voids are named according to the orientation of spheres constituting the voids.
Answer: D
\nOriginal PDF solution page
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Solution: Relative increase in packing efficiency is high when larger voids are occupied.
Answer: A
\nOriginal PDF solution page
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Solution: BCC : 2 3 2 r a = FCC : 2 2 2 r a =
Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Solution: Density depends on mass and size of particles but not the packing efficiency.
Answer: A
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: Packing efficiency of diamond is only 0.34.
Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: \u0394 \u0394 H S = + = + ve ve , Hence for \u2013ve \u0394 G , the temperature should be high.
Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: A \u2192 Q, W; B \u2192 P, U; C \u2192 R, V
\nOriginal PDF solution page
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Solution: Theoretical (Cubic crystals)
Answer: A \u2192 Q, W; B \u2192 P, X; C \u2192 R, Z; D \u2192 S, Y
\nOriginal PDF solution page
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Solution: Informative (Ionic solids)
Answer: A \u2192 P; B \u2192 Q, R; C \u2192 Q, S, T; D \u2192 Q, R
\nOriginal PDF solution page
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Solution: Theoretical (Classification of solids)
Answer: A \u2192 Q; B \u2192 P; C \u2192 S; D \u2192 R
\nOriginal PDF solution page
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Solution: Theoretical (Classification of solids)\n9.37 Solid State HINTS AND EXPLANATIONS
Answer: A \u2192 S; B \u2192 R; C \u2192 P; D \u2192 Q
\nOriginal PDF solution page
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Solution: Theoretical (Classification of solids)
Answer: A \u2192 P; B \u2192 Q, S; C \u2192 R
\nOriginal PDF solution page
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Solution: CaCl : 3 2 a r r = + ( ) + \u2212 CaF and 2 3 4 2 4 : . a r r a r = + ( ) = + \u2212 + Diamond : 3a r = 8 Nacl: + + \u2013 \u2013 = \u221a 2 a 2 a \u221a 2
Answer: A \u2192 R, S; B \u2192 P, Q, R, S; C \u2192 Q
\nOriginal PDF solution page
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Solution: Theoretical (Ionic solids)
Answer: A \u2192 P, S; B \u2192 P, Q; C \u2192 Q; D \u2192 Q, R
\nOriginal PDF solution page
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Solution: Informative (Basic crystal system)
Answer: A \u2192 P, Q, R, S; B \u2192 Q, S; C \u2192 P, R, T; D \u2192 T
\nOriginal PDF solution page
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Solution: Informative (Basic crystal system)
Answer: A \u2192 P, Q, R; B \u2192 T; C \u2192 S; D \u2192 P
\nOriginal PDF solution page
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Solution: CaF and 2 2 2 2 2 : d a d a F F Ca Ca \u2212 \u2212 + \u2212 \u2212 \u2212 = = NaCl Na Na : d a + \u2212 \u2212 = CsCl Cs Cs : d a + + \u2212 = d a T.V. from corner = 3 4 Na O and 2 Na Na O O 2 : d a d d a + + \u2212 \u2212 \u2212 = \u2212 = 2 2 2
Answer: A \u2192 P, T; B \u2192 R; C \u2192 Q; D \u2192 S
\nOriginal PDF solution page
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Solution: SC Nearest Next nearest : , = = a a 2 BCC : = = 3 2 a a FCC : = = 2 2 a a
Answer: 3
\nOriginal PDF solution page
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Solution: Volume cm cm , . . l m d l 3 3 58 5 2 167 27 3 = = = \u21d2 =
Answer: 4
\nOriginal PDF solution page
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Solution: d Z M N V Z Z = \u22c5 \u22c5 \u21d2 = \u00d7 \u00d7 \u00d7 \u00d7 ( ) \u21d2 = \u2212 A 4 72 6 10 0 493 10 4 23 7 3 .
Answer: 4
\nOriginal PDF solution page
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Solution: d d l s CH CH A Z M N V 4 4 ( ) ( ) = = \u22c5 \u22c5 or, 0 5 16 6 10 0 6 10 4 23 7 3 . . = \u00d7 \u00d7 \u00d7 \u00d7 ( ) \u21d2 = \u2212 Z Z
Answer: 4
\nOriginal PDF solution page
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Solution: d Z M N V = \u22c5 \u22c5 A \u21d2 0 92 18 6 10 2 3 4 4 53 10 7 41 10 23 8 2 8 . . . = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa \u2212 \u2212 Z \u2234 Z = 4
Answer: 2
\nOriginal PDF solution page
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Solution: d Z M N V = \u22c5 \u22c5 A \u21d2 2 4 12 6 10 6 3 4 10 2 3 10 23 8 2 8 . = \u00d7 \u00d7 \u00d7 \u00d7 \u22c5 \u00d7 ( ) \u00d7 \u00d7 \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa \u2212 \u2212 Z x \u2234 x 2 200 108 =
Answer: 1
\nOriginal PDF solution page
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Solution: 2 4 a r = \u21d2 2 2 2 1 r a = = nm
Answer: 4
\nOriginal PDF solution page
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Solution: + + \u2013 \u2013 r 2 2 186 2 214 2 400 r = + \u239b \u239d \u239c \u239e \u23a0 \u239f = pm pm\n9.38 Chapter 9 HINTS AND EXPLANATIONS
Answer: 5
\nOriginal PDF solution page
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Solution: d Z M N V = \u22c5 \u22c5 A = \u00d7 \u00d7 \u00d7 \u00d7 ( ) = \u2212 4 58 5 6 10 500 10 3 12 23 10 3 . . gm/cm 3 \u2234 Percentage vacancy = \u2212 \u00d7 = 3 12 2 964 3 12 100 5 . . . %
Answer: 5
\nOriginal PDF solution page
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Solution: (ii), (iv), (v), (vi) , (vii) are true statements.
Answer: 2
\nOriginal PDF solution page
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Solution: d theo 3 gm/cm = \u00d7 \u00d7 \u00d7 \u00d7 ( ) = \u2212 \u2212 4 31 25 1 67 10 500 10 1 67 24 10 3 . . . Now, m theo = 1670 gm per litre m actual = 1607.5 gm per litre \u2234 Moles of metal missing per litre = \u2212 = 1670 1607 5 31 25 2 . .
Answer: 6
\nOriginal PDF solution page
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Solution: Z Z ZnS ZnS but due to defect, = = 4
Answer: 8
\nOriginal PDF solution page
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Solution: 12. Octahedron has eight triangular faces. Hence, truncated octahedron will have eight hexagonal faces.
Answer: 2
\nOriginal PDF solution page
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Solution: BCC: Fraction of edge covered by atom = 2 r a . \u2234 Fraction of edge uncovered = \u2212 = \u2212 \u00d7 = 1 2 1 2 3 4 0 134 r a . From question: 0.134 a = 67 pm \u21d2 a = 500 pm Now, d Z M N V = \u22c5 \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 A 3 gm/cm 2 75 6 10 500 10 2 23 10 3 ( ) ( )
Answer: 2
\nOriginal PDF solution page
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Solution: Let the ideal crystal was having 100 Si-atom, After doping, x Si-atom are missing and y B-atom are doped. Now, ( ) 100 30 11 100 30 88 100 \u2212 \u00d7 + \u00d7 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 x N y N N A A A or, 30 x \u2013 11 y = 360 (1) and ( ) ( . ) . 100 30 11 1 0 001 0 001 3000 30 11 999 \u2212 \u00d7 \u00d7 = \u2212 \u21d2 \u2212 = x N y N x y A A (2) From (1) and (2), y x \u00d7 = 100 2%
Answer: 2
\nOriginal PDF solution page
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Solution: In truncated octahedron, all corner of octahedron become square faces and hence, its number =
Answer: 0143
\nOriginal PDF solution page
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Solution: And, the number of c-atoms per unit cell in diamond =
Answer: 0066
\nOriginal PDF solution page
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Solution: Four-digit Integer Type
Answer: 0045
\nOriginal PDF solution page
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Solution: m d N r \u00d7 = \u00d7 0 74 4 3 3 . A \u03c0 or, 197 19 7 0 74 6 10 4 3 23 3 . . \u00d7 = \u00d7 \u00d7 \u00d7 \u03c0 r \u21d2 r = \u00d7 = \u2212 1 43 10 143 8 . cm pm
Answer: 0072
\nOriginal PDF solution page
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Solution: P.E. of diamond and here silicon is 0.34.
Answer: 0012
\nOriginal PDF solution page
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Solution: d Z M N V = \u22c5 \u22c5 A \u21d2 1 5 10 6 6 10 12 5 8 0 3 0 10 3 23 9 3 . . . . \u00d7 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u00d7 \u00d7 ( ) \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa \u2212 M \u21d2 M = 45 Kg/mol
Answer: 0155
\nOriginal PDF solution page
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Solution: Let the percentage of fayalite be x . V V V olivine fayalite fosterite = + or, 100 3 9 4 2 100 3 3 . . . = + \u2212 x x \u21d2 x = 71.79
Answer: 0018
\nOriginal PDF solution page
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Solution: d Z M N V = \u22c5 \u22c5 A \u21d2 2 14 4 426 18 6 10 1 26 10 23 7 3 . . = \u00d7 + ( ) \u00d7 ( ) \u00d7 \u00d7 ( ) \u2212 x \u21d2 x = 12\n9.39 Solid State HINTS AND EXPLANATIONS
Answer: 0182
\nOriginal PDF solution page
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Solution: For diamond: 3 4 a d = \u00d7 \u2212 C C Now, d Z M N V = \u22c5 \u22c5 A \u21d2 2 3 8 12 6 10 4 3 23 3 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 d C C \u2234 d C C cm pm \u2212 \u2212 = \u00d7 = 1 55 10 155 8 .
Answer: 6041
\nOriginal PDF solution page
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Solution: The maximum backing efficiency of identical spheres in 2D is 0.90. Hence, 40 0 90 10 2 2 2 ( ) \u00d7 = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f . N \u03c0 \u21d2 N = 18
Answer: 0067
\nOriginal PDF solution page
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Solution: 4 2 520 . r Cl \u2212 = \u00d7 \u21d2 r Cl pm \u2212 = 182
Answer: 0059
\nOriginal PDF solution page
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Solution: Z Ti = \u00d7 = 1 2 1 1 2 Z O = 1 (assume) \u2234 Formula = Ti O T O i 1 2 1 2 / \u2245 Now, Ti = + \u00d7 = 48 48 32 100 60% and oxidation state of Ti = +4
Answer: 0125
\nOriginal PDF solution page
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Solution: For one litre crystal, m m m m initial Bremoved Cadded final \u2212 + = or, 4800 30 1 15 4795 \u2212 \u00d7 + \u00d7 = x \u21d2 x = 2 3 \u2234 Percentage of C-atoms which replaced B-atoms = \u00d7 = x 1 100 67%
Answer: 0073
\nOriginal PDF solution page
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Solution: Percentage of body diagonal covered = + + ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u00d7 \u2212 + \u2212 2 4 3 2 100 r r r r B C A B % = + \u00d7 + ( ) \u00d7 = \u2212 \u2212 \u2212 \u2212 2 4 0 225 2 3 0 414 100 59 2 r r r r B B B B . . % . %
Answer: 4000
\nOriginal PDF solution page
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Solution: Out of 8 Fe 2+ in original crystal, 1 is missing. \u2234 Percentage of cation vacancy = \u00d7 = 1 8 100 12 5 . %
Answer: 1000
\nOriginal PDF solution page
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Solution: Percentage occupied space = \u00d7 = \u00d7 \u00d7 \u00d7 ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 V V particles solid 100 6 10 4 3 1 54 10 40 4 1 23 8 3 % . / \u03c0 \u03c0 0 00 27 % % = \u2234 Empty space = 73 %
![\"Surface](\"./assets/chapter-packs/surface-chemistry/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: Greater the specific surface area of adsorbent, greater will be the extent of adsorption.
Answer: B
\nOriginal PDF solution page
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Solution: Adsorption decreases the surface energy.
Answer: A
\nOriginal PDF solution page
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Solution: x m K P x m K n P n = \u21d2 = + \u22c5 . log log log 1 1 From question, log K = 0.3010 = log 2 \u21d2 K = 2 And 1 45 1 1 n n = \u00b0 = \u21d2 = tan \u2234 x m P = \u00d7 = \u00d7 = 2 2 0 2 0 4 . .
Answer: B
\nOriginal PDF solution page
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Solution: K A e E RT a = \u2212 . / \u2234 ln K K E R T T a 2 1 1 2 1 1 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, ln 10 cal/mol = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = E R E a a 1 600 1 1000 6900
Answer: A
\nOriginal PDF solution page
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Solution: For a particular combination of adsorbent, adsorbate and temperature, only one value of \u2018 n \u2019 is permissible.
Answer: B
\nOriginal PDF solution page
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Solution: x m K P n = \u22c5 1 \u21d2 0 2 4 1 . ( ) = \u00d7 K n (1) 0 5 25 1 . ( ) = \u00d7 K n (2) 0 8 64 1 . ( ) = \u00d7 K n (3) From (1), (2) and (3), K n = = 1 10 2 and \u2234 x m K n \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 = required ( ) . 36 0 6 1 Hence, moles of N 2 adsorbed per gm of iron = = 0 6 28 3 140 .
Answer: D
\nOriginal PDF solution page
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Solution: r r 1 2 < \u21d2 A is the catalyst. r r r r r 3 1 1 4 5 < \u21d2 = = \u21d2 B is the catalyst. C and D are not catalysts. . r r r r r r C 1 7 2 1 2 6 < < \u21d2 < < \u21d2 D is catalytic poison. is catalytic prom motor.
Answer: D
\nOriginal PDF solution page
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Solution: A catalyst always involve in the reaction.
Answer: C
\nOriginal PDF solution page
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Solution: Enzymes are specific.
Answer: A
\nOriginal PDF solution page
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Solution: A catalyst does not initiate the reaction.
Answer: A
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: Catalyst does not alter the equilibrium position.
Answer: C
\nOriginal PDF solution page
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Solution: Homogeneous catalysis, because the physical states of both reactant and catalyst is aqueous (liquid).
Answer: D
\nOriginal PDF solution page
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Solution: Activation energy is decreased.
Answer: A
\nOriginal PDF solution page
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Solution: Catalyst lowers the activation energy.
Answer: D
\nOriginal PDF solution page
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Solution: Catalysis occurs through chemisorption.
Answer: C
\nOriginal PDF solution page
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Solution: Viscosity is higher and surface tension is smaller than water.
Answer: A
\nOriginal PDF solution page
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Solution: K , as it reacts vigorously in water.
Answer: A
\nOriginal PDF solution page
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Solution: SnCl 4 formed by reaction will adsorb some common Cl \u2013 ions.
Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: True solution or suspension does not show Tyndall eff ect.
Answer: D
\nOriginal PDF solution page
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Solution: Blood, clay and smoke are negative sol. In strong acidic solution, gelatin adsorbs some H + ions and become positive.
Answer: B
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Solution: Informative EXERCISE II (JEE ADVANCED)\n12.24 Chapter 12 HINTS AND EXPLANATIONS
Answer: A
\nOriginal PDF solution page
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Solution: Volume of metal used = \u00d7 = \u2212 \u2212 1 9 10 19 10 4 5 . cm 3 \u2234 N \u00d7 \u00d7 \u00d7 = \u2212 \u2212 4 3 10 10 10 7 3 5 3 ( ) cm cm \u21d2 N = \u00d7 2 39 10 12 . Hence, number of particles per cm 3 = \u00d7 = \u00d7 2 39 10 1000 2 39 10 12 9 . .
Answer: A
\nOriginal PDF solution page
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Solution: Larger the carbon chain, normally smaller is CMC.
Answer: C
\nOriginal PDF solution page
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Solution: Sulphide sol have negative charge on colloidal particles
Answer: B
\nOriginal PDF solution page
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Solution: Adsorption is physisorption.
Answer: B
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Solution: Entropy decreased in adsorption.
Answer: A, B, C, D
\nOriginal PDF solution page
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Answer: A, C
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: \u0394 H can never be equal to \u0394 S.
Answer: A, D
\nOriginal PDF solution page
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Solution: (a) Chemisorption does not change into physisorption at higher pressure. (b) CO or CO 2 gases leave the surfaces.
Answer: A, B, D
\nOriginal PDF solution page
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Answer: A, B, C, D
\nOriginal PDF solution page
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Answer: A, B, C
\nOriginal PDF solution page
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Solution: In Lindlar\u2019s catalyst, catalytic poison is used
Answer: A, C
\nOriginal PDF solution page
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Answer: B, D
\nOriginal PDF solution page
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Solution: All catalytic reaction is multistep reaction
Answer: A, B, C
\nOriginal PDF solution page
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Answer: A, C, D
\nOriginal PDF solution page
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Solution: (a) K K e A e A e e E RT E RT E E RT a a a a cat uncat = = = \u2032 \u2032 \u2032 \u2212 20 . . / / ( )/ \u2234 20 2 2 1000 300 = \u2212 = \u2212 \u00d7 \u2032 E E RT E a a a Kcal \u21d2 E a = 14 Kcal/mol (b), (c) Reaction: 2H 2 O 2 (aq) \u2192 2H 2 O(l)+O 2 (g) is fi rst order. (d) Rate of uncatalysed reaction increases to greater extent on increasing temperature because its activation energy is high.
Answer: C, D
\nOriginal PDF solution page
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Solution: Sol particles are restricted to move. Solvent particles move in opposite direction to the expected movement of sol particles. Fe(OH) 3 sol is positively charged.
Answer: A, B, C, D
\nOriginal PDF solution page
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Answer: B, C
\nOriginal PDF solution page
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Solution: Electrophoresis and electro-osmosis are the experimental methods to determine charge on colloidal particles.
Answer: A, C
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Solution: Below CMC, the solution is true solution.
Answer: A, B, C
\nOriginal PDF solution page
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Solution: PO SO Cl 4 3 4 2 \u2212 \u2212 \u2212 > > \u21d2 Sol particles are positively charged.
Answer: A, B, C
\nOriginal PDF solution page
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Solution: RCOONa RCOO Na \u001c \u2212 + + As true solution, one mole of RCOONa will become two moles in solution. But, as micelle formation starts, the total number of particles start decreasing due to association.
Answer: C, D
\nOriginal PDF solution page
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Solution: Due to excess Ag+, sol particles will be positively charged.
Answer: C
\nOriginal PDF solution page
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Solution: (a) It is due to sharp decrease in number of ions. (b) Tyndall effect is better shown by lyophobic colloid. (c) Colloidal solutions have lower value of colligative properties. (d) Larger the carbon chain, normally lower CMC value.
Answer: A, B, D
\nOriginal PDF solution page
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Solution: (a) Basic dye is positively charged and hence, Fe(CN) HPO 6 4 3 2 \u2212 \u2212 > (c) Slope should not change in Freundlich\u2019s isotherm.
Answer: A, C
\nOriginal PDF solution page
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Solution: Tyndall effect is shown by colloids.
Answer: A, D
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: Charge : Mg 2+ (2 unit) > Cl \u2013 (1 unit) Hence, better coagulation for negatively charged gold sol.\n12.25 Surface Chemistry HINTS AND EXPLANATIONS
Comprehension 1 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
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+ "D"
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+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: Polarizability is maximum in Xe.
Comprehension 1 - subquestion 2
",
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+ "identifier": "A",
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+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
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+ {
+ "identifier": "D",
+ "content": ""
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+ "C"
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+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: CO is polar and hence, more preferential adsorption.
Comprehension 1 - subquestion 3
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
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+ ],
+ "correct_options": [
+ "B"
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+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Adsorption decreases on increasing temperature. Comprehension II
Comprehension 2 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
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+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
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+ {
+ "identifier": "D",
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+ "A"
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+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: In case of concentrated KCl, KCl adsorbs on blood charcoal surface, but in case of dilute KCl, blood charcoal dissolves in KCl solution.
Comprehension 2 - subquestion 2
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
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+ "C"
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+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: Greater critical temperature, greater the extent of adsorption.
Comprehension 2 - subquestion 3
",
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+ "identifier": "A",
+ "content": ""
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+ "identifier": "B",
+ "content": ""
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+ {
+ "identifier": "C",
+ "content": ""
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+ {
+ "identifier": "D",
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+ "A"
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+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: Adsorption is always exothermic. Comprehension III
Comprehension 3 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
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+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
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+ "D"
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+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: Initial surface area, A 1 2 2 6 2 24 = \u00d7 = ( ) cm cm Final volume of each cube = cm 3 8 10 12 \u2234 Final side length of each cube = \u00d7 ( ) 8 10 12 1 3 cm 3 / = \u00d7 \u2212 2 10 4 cm Hence, final surface area of each cube, A 2 = 6 \u00d7 (2 \u00d7 10 \u22124 cm ) 2 = 24 \u00d7 10 \u22128 cm 2 \u2234 Final total surface area Initial surface area = \u00d7 \u00d7 \u2212 24 10 10 2 8 12 4 4 10 4 =
Comprehension 3 - subquestion 2
",
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+ {
+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
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+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
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Solution: Number of H 2 molecules = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 2 0 112 0 0821 546 6 10 3 10 23 21 . . \u2234 Specific surface area = \u00d7 \u00d7 \u00d7 \u2212 3 10 0 4 10 5 21 7 2 . ( ) cm gm = \u00d7 2 4 10 6 . cm /gm 2 Comprehension IV
Comprehension 3 - subquestion 3
",
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+ "identifier": "A",
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+ "identifier": "B",
+ "content": ""
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+ {
+ "identifier": "C",
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+ "A"
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+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: log log log x m K n P \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u22c5 1 Slope = = \u21d2 = 1 0 25 4 n n . and for x -intercept, log x m \u239b \u239d \u239c \u239e \u23a0 \u239f = 0 \u21d2 log log K n P = \u2212 \u22c5 1 = \u2212 \u00d7 \u2212 = 1 4 4 1 0 ( ) . \u2234 K = 10
Comprehension 4 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
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+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
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+ "identifier": "D",
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+ ],
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+ "A"
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+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: x m = \u00d7 = 10 16 20 1 4 ( ) / \u21d2 x = \u00d7 = 20 10 200 gm
Comprehension 4 - subquestion 2
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
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+ "identifier": "D",
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+ "B"
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+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: 810 1 0 10 1 4 . ( ) / = \u00d7 P \u21d2 P = 3 atm\n12.26 Chapter 12 HINTS AND EXPLANATIONS Comprehension V
Comprehension 4 - subquestion 3
",
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+ "B"
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+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: Positively charged due to adsorption of Ag + ions.
Comprehension 5 - subquestion 1
",
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+ },
+ {
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+ ],
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+ "B"
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+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Comprehension 5 - subquestion 2
",
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+ "identifier": "A",
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+ "identifier": "B",
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+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
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+ "C"
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+ "explanation": "Answer: C
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Solution: Informative Comprehension VI
Comprehension 5 - subquestion 3
",
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+ "identifier": "C",
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+ "D"
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+ "explanation": "Answer: D
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Solution: Positively charged due to adsorption of Fe 3+ ions.
Comprehension 6 - subquestion 1
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
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+ ],
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+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
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Solution: AgI Ag NO fixed la , , - + \u2193 \u2193 3 y yer diffused layer
Comprehension 6 - subquestion 2
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
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+ "identifier": "C",
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+ },
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+ "identifier": "D",
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+ ],
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+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
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Solution: S 2\u2013 ions get adsorbed. Comprehension VII
Comprehension 6 - subquestion 3
",
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+ "identifier": "C",
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+ "B"
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+ "explanation": "Answer: B
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Solution: Gold number = 0.025 \u00d7 1000 = 25
Comprehension 7 - subquestion 1
",
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+ "identifier": "A",
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+ "identifier": "B",
+ "content": ""
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+ {
+ "identifier": "C",
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+ },
+ {
+ "identifier": "D",
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+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Comprehension 7 - subquestion 2
",
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+ "identifier": "A",
+ "content": ""
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+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
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+ "C"
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+ "answer": null,
+ "explanation": "Answer: C
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Answer: A
\nOriginal PDF solution page
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Solution: Colour become less intense due to adsorption.
Answer: B
\nOriginal PDF solution page
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Solution: Surface particles have higher energy due to unbalanced forces.
Answer: D
\nOriginal PDF solution page
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Solution: Word 'always' is not suitable because chemisorption increased with increase in temperature.
Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Answer: B
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Solution: Cellulose nitrate sol is lyophilic.
Answer: B
\nOriginal PDF solution page
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Solution: Scattering is not related to speed of particles.
Answer: B
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Answer: A
\nOriginal PDF solution page
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Solution: Colloidal particles are negatively charged due to adsorption of I - ions and hence, it moves towards anode.
Answer: A
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Answer: C
\nOriginal PDF solution page
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Solution: Peptization occurs due to adsorption of common ion.
Answer: A
\nOriginal PDF solution page
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Solution: Natural colloids are normally lyophilic.
Answer: B
\nOriginal PDF solution page
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Answer: A \u2192 Q, U; B \u2192 P, W; C \u2192 P, V; D \u2192 R, X
\nOriginal PDF solution page
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Answer: A \u2192 R; B \u2192 S; C \u2192 P; D \u2192 Q
\nOriginal PDF solution page
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Answer: A \u2192 Q; B \u2192 P; C \u2192 Q, R; D \u2192 P, S
\nOriginal PDF solution page
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Answer: A \u2192 Q; B \u2192 S; C \u2192 P; D \u2192 R
\nOriginal PDF solution page
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Solution: Informative\n12.27 Surface Chemistry HINTS AND EXPLANATIONS
Answer: 4
\nOriginal PDF solution page
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Solution: 6 48 0 01 10 162 10 3 3 . ( . ) = \u00d7 \u00d7 \u00d7 \u2212 n \u21d2 n = 4
Answer: 3
\nOriginal PDF solution page
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Solution: log log log x m K n P = + \u22c5 1 From the given graph, log K = 1.0 \u21d2 K = 10 and 1 0 25 n = . \u21d2 n = 4 Now, x m K P n = . 1 \u21d2 x 1 0 10 8 1 10 3 1 4 . ( . ) / = \u00d7 \u00d7 \u2212 \u21d2 x = 3 gm
Answer: 4
\nOriginal PDF solution page
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Solution: t k A e s e av E RT a = = = \u00d7 \u00d7 = \u2212 \u2212 \u2212 \u00d7 \u00d7 1 1 1 1 25 10 4 8 1 16 10 2 400 3 . ( . ) sec / /
Answer: 6
\nOriginal PDF solution page
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Solution: Soap solution of sodium palmitate, gold sol, silicic acid sol, acidic dye, metal sulphide sol, sol of AgCl by excess KCl in AgNO 3 .
Answer: 6
\nOriginal PDF solution page
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Solution: ln P P R T T ads 1 2 1 2 1 1 = \u2212 \uf8eb \uf8ed \uf8ec \uf8ec \uf8ec \uf8ec \uf8f6 \uf8f8 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \u0394 H or, ln H 1 6 32 1 200 1 250 . = \u2212 \uf8eb \uf8ed \uf8ec \uf8ec \uf8ec \uf8f6 \uf8f8 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \u0394 ads R \u21d2 \u2206 H ads = 6000 cal/mol
Answer: 5
\nOriginal PDF solution page
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Solution: Number of CH 3 COOH molecules adsorbed = \u00d7 \u2212 \u00d7 \u00d7 = \u00d7 100 0 5 0 49 1000 6 10 6 10 23 20 ( . . ) \u2234 Surface area of each molecule = \u00d7 \u00d7 = \u00d7 \u2212 3 10 6 10 5 10 2 20 19 m 2
Answer: 2
\nOriginal PDF solution page
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Solution: Number of N 2 molecules = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 0 001 2 46 10 0 082 300 6 023 10 6 023 10 3 23 16 . . . . . \u2234 Number of active sites per molecule = \u00d7 \u00d7 \u00d7 \u00d7 = 1000 6 023 10 20 100 6 023 10 2 14 16 . .
Answer: 9
\nOriginal PDF solution page
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Solution: Number of N 2 molecules absorbed = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 2 24 10 22 4 6 10 6 10 3 23 19 . . \u2234 Specific surface area = \u00d7 \u00d7 \u00d7 = \u2212 6 10 0 15 10 9 19 9 2 . ( )
Answer: 2
\nOriginal PDF solution page
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Solution: Colloid is a heterogeneous system \u21d2 min = 2 phases
Answer: 9
\nOriginal PDF solution page
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Solution: ln 20 1 600 1 1000 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E R a \u21d2 E a = 9000 cal/mol. Four-digit Integer Type
Answer: 0600
\nOriginal PDF solution page
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Solution: Mass of NaCl used = \u00d7 \u00d7 = \u00d7 ( ) 585 1 2 1 100 5 85 1 2 . / . . gm Moles of NaCl used = \u00d7 = 5 85 1 2 58 5 0 12 . . . . \u2234 Coagulation value = \u00d7 = 0 12 10 200 1000 600 3 . / millimole/litre
Answer: 0020
\nOriginal PDF solution page
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Solution: Number of palmitic acid molecules needed = \u00d7 = \u00d7 \u2212 480 0 2 10 2 4 10 7 2 17 cm cm 2 . ( ) . Moles of palamitic acid molecules = \u00d7 \u00d7 = \u00d7 \u2212 2 4 10 6 10 4 10 17 23 7 . \u2234 Volume of solution needed = \u00d7 \u00d7 = \u00d7 = \u2212 \u2212 1 5 12 256 4 10 2 10 20 7 5 3 dm dm mm 3 3 . /\n12.28 Chapter 12 HINTS AND EXPLANATIONS
Answer: 0015
\nOriginal PDF solution page
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Solution: The radius of hydrogen molecule = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f m d 3 4 1 3 \u03c0 / = \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239c \u239c \u239e \u23a0 \u239f \u239f \u239f = \u00d7 \u2212 2 6 10 0 16 3 4 2 5 10 23 1 3 8 . . / \u03c0 \u03c0 cm Number of hydrogen molecules at the surface per gm Cu = \u00d7 \u00d7 = \u00d7 224 22400 6 10 25 2 4 10 23 20 / . \u03c0 \u03c0 \u2234 Specific surface area of Cu = cm /gm m /gm 2 2 2 4 10 2 5 10 150000 15 20 2 8 . ( . ) \u00d7 \u00d7 \u00d7 \u00d7 = = \u2212 \u03c0 \u03c0
Answer: 0014
\nOriginal PDF solution page
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Solution: t k av = 1 Now, ln ln k k t t E R T T a 2 1 1 2 1 2 1 1 = = = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, ln . . 0 36 0 72 1 2500 1 2000 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E R a \u21d2 E a = 14000 cal/mol
Answer: 0010
\nOriginal PDF solution page
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Solution: t t K K A e A e Fe 1 2 1 2 20 10 2 600 8 3 / / / . . ( ) ( ) = = \u2212 \u00d7 \u00d7 \u2212 \u00d7 Fe charcoal charcoal 1 10 2 600 10 3 1 / \u00d7 = e
Answer: 0025
\nOriginal PDF solution page
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Solution: Number of adsorbate molecules = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 0 10 10 0 25 10 6 10 2 4 10 3 3 23 7 . . . \u2234 Eff ective surface area = \u00d7 = \u00d7 \u2212 0 06 2 4 10 25 10 17 20 2 . . m
Answer: 0091
\nOriginal PDF solution page
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Solution: Moles of gas adsorbed per gm of charcoal = \u2212 \u00d7 \u00d7 \u00d7 \u00d7 ( ) . 700 400 1 52 760 300 6 R Volume of gas adsorbed per gm of charcoal (at 0\u00b0C and 1 atm) = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 300 1 52 760 300 6 273 1 . R R = 0.091 litre
Answer: 0720
\nOriginal PDF solution page
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Solution: Specific surface area of silica gel = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 168 22400 6 10 0 16 10 720 23 9 2 . ( ) m /gm 2
Answer: 0400
\nOriginal PDF solution page
Open page 617 in PDFExtracted text
Solution: ln P P R T T 1 2 1 1 1 1 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u0394 \u03b8 H or, ln . . . . 0 4 59 2 16 628 10 8 314 1 200 1 3 = \u2212 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f T \u21d2 T = 400 K
Answer: 4000
\nOriginal PDF solution page
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Solution: r K K E S K K K S K K E S K K S = + + + \u2212 \u2212 1 2 0 1 2 1 1 2 0 1 1 [ ][ ] [ ] [ ][ ] [ ] \u001b For r max , K S K 1 1 [ ] \u001a \u2212 \u2234 r max = = = K K E S K S K E 1 2 0 1 2 0 0 02 [ ][ ] [ ] [ ] . M From question, K E K K E S K K S 2 0 1 2 0 1 1 2 [ ] [ ][ ] [ ] = + \u2212 \u21d2 K K S K S \u2212 + = 1 1 1 2 [ ] [ ] \u2234 K K S 1 1 3 3 6 3 1 1 250 250 10 4000 \u2212 \u2212 = = = \u00d7 = [ ] mg dm dm kg dm kg
![\"Thermochemistry](\"./assets/chapter-packs/thermochemistry/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
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Solution: N O NO Brown 2 4 2 2 \u001f On heating, colour deepens means reaction is endothermic \u0394 H ve = + ( ) .
Answer: B
\nOriginal PDF solution page
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Solution: As the process is endothermic but no heat is absorbed from surrounding, the temperature of the system will decrease, As initial temperature is used, the calculated mole will be lower.
Answer: B
\nOriginal PDF solution page
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Solution: \u0394 \u0394 \u0394 \u0394 \u0394 H H H H 2 1 2 1 2 1 3 2 0 \u2212 \u2212 = ( ) = ( ) \u2212 + ( ) = \u21d2 = T T C x x x P
Answer: D
\nOriginal PDF solution page
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Solution: I 2 (s) \u2192 I 2 (g); \u0394 \u0394 H cal/gm at K H at K 1 1 2 2 24 473 523 = = = = T T ? \u0394 \u0394 \u0394 H H H 2 1 2 1 2 2 2 24 523 473 0 055 0 031 \u2212 \u2212 = ( ) \u2212 ( ) \u21d2 \u2212 \u2212 = \u2212 T T C I g C I s P P , , . . \u2234 \u0394 H c al/gm 2 25 2 = .
Answer: B
\nOriginal PDF solution page
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Solution: For direct measurement, reaction must occur directly in the conditions to measure heat.
Answer: A
\nOriginal PDF solution page
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Solution: Greater the mass per cent of hydrogen, greater is the calorific value.
Answer: D
\nOriginal PDF solution page
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Solution: For the reaction: \u0394 \u0394 \u0394 U H RT kJ = \u2212 \u22c5 = \u2212 ( ) \u2212 \u2212 ( ) \u00d7 \u00d7 = \u2212 n g 72 3 1 8 314 1000 298 69 8 . . . As HCl is limiting reagent, for the given amount, \u0394 U kJ = \u00d7 \u2212 ( ) = \u2212 2 69 8 139 6 . .\n5.36 Chapter 5 HINTS AND EXPLANATIONS
Answer: C
\nOriginal PDF solution page
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Solution: HAuBr 4 + 4HCl \u2192 HAuCl 4 + 4HBr; \u0394 H = (\u201328) \u2013 (\u201336.8) = 8.8 kcal \u2234 Percentage reaction = \u00d7 = 0 44 8 8 100 5 . . %
Answer: A
\nOriginal PDF solution page
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Solution: (a) C + O CO; 1 2 9 0 4 5 2 \u2192 . . \u0394 H = \u201375 kcal Heat evolved = 75 \u00d7 9 = 675 kcal (b) C + O CO 2 2 2 2 \u2192 ; \u0394 H = \u201395 kcal Heat evolved = 95 \u00d7 2 = 190 kcal (c) 4C O CO 3CO + \u2192 + 3 5 2 2 . Heat evolved = 75 \u00d7 1 + 95 \u00d7 3 = 360 kcal (d) C + O CO 2 2 2 5 2 5 \u2192 . . \u0394 H = \u201395 kcal Heat evolved = 95 \u00d7 2.5 = 237.5 kcal
Answer: A
\nOriginal PDF solution page
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Solution: CH3 NO2 NO2 O2N l O g CO g H O l N g ( ) + ( ) \u2192 ( ) + ( ) + ( ) 21 4 7 5 2 3 2 2 2 2 2 \u0394 H kJ/mol k = \u00d7 \u2212 ( ) + \u00d7 \u2212 ( ) \u2212 ( ) = \u2212 = \u2212 \u00d7 7 395 5 2 285 65 3542 5 3542 5 227 1 816 . . . J J/mol kJ/mol MJ/L = \u2212 = \u2212 28 34 28 34 . .
Answer: D
\nOriginal PDF solution page
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Solution: For H A 2 , enthalpy of ionization = \u00d7 \u2212 ( ) = 2 13 5 13 1 . kcal/mol For B OH 2 ( ) , enthalpy of ionization = \u00d7 \u2212 ( ) = 2 13 5 10 7 . kcal/mol \u2234 Required \u0394 = \u00d7 \u2212 \u2212 = \u2212 H kcal 2 13 5 1 7 19 .
Answer: D
\nOriginal PDF solution page
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Solution: If BaSO 4 were water soluble, then \u0394 = \u00d7 \u2212 ( ) = \u2212 H kJ expected 2 57 114
Answer: B
\nOriginal PDF solution page
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Solution: Required \u0394 = \u2212 \u2212 \u00d7 ( ) = \u2212 H cal 13700 400 0 9 13340 .
Answer: A
\nOriginal PDF solution page
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Solution: Aerobic oxidation results release of energy and hence, it is biologically benefical by (2880 + 2530 = 5410 kJ/mol)
Answer: D
\nOriginal PDF solution page
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Solution: (a) Si H g H g SiH g H kcal 2 6 2 4 ( ) + ( ) \u2192 ( ) \u0394 = \u2212 2 11 7 ; . (b) SiH g SiH g H g H kcal 4 2 2 ( ) \u2192 ( ) + ( ) = + ; . \u0394 239 7 (c) 2Si s H g Si H g H kcal 2 ( ) + ( ) \u2192 ( ) = + 3 80 3 2 6 ; . \u0394 Required thermochemical equation is Si s H g SiH g 2 2 ( ) + ( ) \u2192 ( ) From (b) a) (c) H kcal/mol + + = + 1 2 1 2 274 ( : \u0394
Answer: B
\nOriginal PDF solution page
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Solution: Required thermochemical equation is Dy s Cl g DyCl s 2 3 ( ) + ( ) \u2192 ( ) 3 2 From (ii) + 3 \u00d7 (iii) \u2013 (i), we get: \u0394 H kJ/mol = \u2212 ( ) + \u2212 ( ) \u2212 \u2212 ( ) = \u2212 699 43 3 158 31 180 06 994 3 . . . .
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 \u2212 ( ) + \u2212 ( ) + \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 ( ) + \u2212 H= 188 84 22 05 2 22 1 2 17 63 70 97 8 5 . . . . . . 6 6 0 2 68 32 74 18 ( ) + + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 . . kcal
Answer: C
\nOriginal PDF solution page
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Solution: H aq OH aq H O l 2 + \u2212 ( ) + ( ) \u2192 ( ) \u0394 = \u0394 \u2212 \u0394 + \u0394 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ( ) ( ) ( ) + \u2212 H H H H f H O l f H aq f OH aq 2 or, \u2212 = \u2212 ( ) \u2212 + \u0394 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 ( ) 57 32 285 84 0 . . f OH aq H \u2234 \u0394 = \u2212 \u2212 ( ) f OH aq H kJ/mol 228 52 .
Answer: C
\nOriginal PDF solution page
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Solution: CH CH COOH l O g CO g H O l 3 2 2 2 2 7 2 3 3 ( ) + ( ) \u2192 ( ) + ( ) \u0394 = \u00d7 \u0394 + \u00d7 \u0394 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u0394 ( ) ( ) ( ) C CH CH COOH l f CO g f H O l f CH CH COOH H H H H 3 2 2 2 3 2 3 3 l l f O g H ( ) ( ) + \u00d7 \u0394 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 7 2 2 or, 3 3 94 3 68 3 2 3 2 \u00d7 \u0394 = \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u0394 \u23a1 \u23a3 \u23a4 ( ) ( ) f CH CH COOH f CH CH COOH H H +O l l \u23a6 \u23a6 \u2234 \u0394 = \u2212 ( ) f CH CH COOH H kcal/mol 3 2 121 5 l .\n5.37 Thermochemistry HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: phOH(solution II) \u2192 phOH(solution I) \u0394 = \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 H kcal/mol 0 02 0 47 94 0 03 1 410 94 2 . . . .
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 = \u2212 ( ) \u2212 \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 [ ] = + H kcal/mol required 14 7 2 7 13 75 1 75 . . . .
Answer: A
\nOriginal PDF solution page
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Solution: 2 1 2 197 2 65 2 3 FeO O Fe O H 2 + \u2192 = \u2212 ( ) \u2212 \u2212 ( ) ; \u0394 = \u2212 67 kcal Initial Final 2 2 2 a a a x a x \u2212 + 2 2 1 2 3 5 a x a x x a \u2212 + = \u21d2 = and heat released = 67 x kcal \u2234 Heat released per mole of initial mixture = = 67 3 13 4 x a . kcal
Answer: B
\nOriginal PDF solution page
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Solution: 3C H OH 25H O 3C H OH 25H O 2 5 2 2 5 2 + \u2192 ( ) ; \u0394 \u0394 H kcal H cal theo exp = \u2212 ( ) + \u2212 ( ) = \u2212 = \u2212 ( ) = \u2212 1120 2 1760 4640 3 1650 4950 As experimentally, more heat is released means the mixing is exothermic by (4950 \u2013 4640) = 310 cal.
Answer: D
\nOriginal PDF solution page
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Solution: Heat absorbed in solubility = Heat released from solution = \u0394 = + ( ) \u00d7 \u00d7 = m.s. T J 200 25 4 2 3 2835 . \u2234 \u0394 H J = + \u00d7 = + 2835 7 45 74 5 28350 . .
Answer: A
\nOriginal PDF solution page
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Solution: Heat released by reaction = Heat gained by ice = = \u00d7 = m.L cal 0 2 80 16 . \u0394 \u2212 = \u2212 \u00d7 \u2212 H= cal 16 10 16 10 3 3
Answer: B
\nOriginal PDF solution page
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Solution: let C 2 H 6 = x L, then CH 4 = (4 \u2013 x )L Volume of CO 2 produced, 2 x + (4 \u2013 x ) = 6 \u21d2 x = 2 \u2234 Total heat evolved = \u2212 \u00d7 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 = 1 2 1573 1 2 890 1 0 0821 300 50 . kJ
Answer: A
\nOriginal PDF solution page
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Solution: C O CO H cal + \u2192 \u0394 = \u2212 \u00d7 = \u2212 2 2 2400 0 3 12 96000 ; . C O CO H cal + \u2192 \u0394 = \u2212 \u00d7 = \u2212 1 2 1400 0 6 12 28000 2 ; . Now, CO O CO H cal + \u2192 \u0394 = \u2212 \u2212 ( ) \u2212 \u2212 ( ) = \u2212 1 2 96000 28000 68000 2 2 ; \u2234 Heat produced cal = \u00d7 = 68000 28 0 7 1700 .
Answer: B
\nOriginal PDF solution page
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Solution: Heat liberated from propane = Heat absorbed by water or, n n \u00d7 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u21d2 = 500 10 40 100 160 10 1 50 40 3 3
Answer: B
\nOriginal PDF solution page
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Solution: q = v . i . t = 15 \u00d7 0.125 \u00d7 (14 \u00d7 60) J = 1575 J \u2234 \u0394 = \u00d7 = H J/mol 1575 0 1 1 15750 .
Answer: A
\nOriginal PDF solution page
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Solution: H g O g H O g H 240 kJ 2 2 2 1 1 2 ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; 2H g O g H O H 672 kJ ( ) + ( ) \u2192 ( ) \u0394 = \u2212 + ( ) = \u2212 1 2 240 432 2 2 2 g ; \u2234 \u0394 \u0394 = \u2212 \u2212 = H H 2 1 672 240 2 8 .
Answer: A
\nOriginal PDF solution page
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Solution: \u0394 = = \u00d7 \u00d7 \u00d7 ( ) = \u00d7 \u2212 m E C kg 2 3 8 2 12 103 10 4 2 3 10 4 8 10 . .
Answer: B
\nOriginal PDF solution page
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Solution: \u2212 = \u2212 ( ) + + ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 ( ) + \u2212 ( ) + + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 12 75 3 9 11 8 5 17 5 . . . x \u2234 X = \u2212 22 1 . kcal/mol
Answer: A
\nOriginal PDF solution page
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Solution: Dulong and petit\u2019s law : Atomic mass \u00d7 Specific heat \u2245 6.4 for greater temperature rise, heat lost should be high.\n5.38 Chapter 5 HINTS AND EXPLANATIONS
Answer: B
\nOriginal PDF solution page
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Solution: g O g CO g H O l ( ) + ( ) \u2192 ( ) + ( ) 9 2 3 3 2 2 2 \u0394 = \u00d7 \u0394 + \u00d7 \u0394 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u0394 ( ) ( ) C cyclopropane f CO g f H O l f cyclopropan H H H H 3 3 2 2 e e f O H g + \u00d7 \u0394 ( ) \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 9 2 2 = \u00d7 \u2212 ( ) + \u00d7 \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 ( ) + ( ) + \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 3 394 3 286 33 20 9 2 0 { } = \u2212 2093 kJ/mol
Answer: B
\nOriginal PDF solution page
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Solution: w = \u2212 = \u2212 = \u2212 \u00d7 \u00d7 = \u2212 P.V n RT cal H 2 1 5 2 298 894 .
Answer: C
\nOriginal PDF solution page
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Solution: w = \u2212 = \u2212 \u00d7 \u00d7 = \u2212 nRT cal 1 2 353 706 \u2234 \u0394 = + = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = U kcal/mol q w 7 4 706 1000 6 694 . .
Answer: A
\nOriginal PDF solution page
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Solution: C H g +5O g 3CO g +4H O l 3 8 2 2 2 ( ) ( ) \u2192 ( ) ( ) \u0394 = \u00d7 \u0394 + \u00d7 \u0394 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u0394 + \u00d7 \u0394 ( ) ( ) ( ) ( ) C C H f CO g f H O l f C H g f O g H H H H H 3 3 2 2 3 3 2 3 4 5 g ( ( ) \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 ( ) = \u2212 3 393 5 4 285 8 103 8 2219 9 . . . . kJ \u0394 \u0394 \u0394 \u0394 \u0394 r required C C H g C CH g C C H g C H g H H H H H 2 6 4 3 8 2 = \u2212 + \u23a1 \u23a3 \u23a4 \u23a6 + + \u23a1 ( ) ( ) ( ) ( ) \u23a3 \u23a3 \u23a4 \u23a6 = \u2212 \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 + \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 1560 0 890 0 2219 9 285 8 55 7 . . . . . kJ J
Answer: B
\nOriginal PDF solution page
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Solution: The required thermochemical equation is K s Cl g KCl s ; H 2 ( ) + ( ) \u2192 ( ) = 1 2 \u0394 ? From (iv) + (iii) \u2013 (v) + (i) \u2013 (ii): we get, \u0394 H Kcal = \u2212 ( ) + \u2212 ( ) \u2212 ( ) + \u2212 ( ) \u2212 \u2212 ( ) = \u2212 116 5 39 3 4 4 13 7 68 4 105 5 . . . . . .
Answer: A
\nOriginal PDF solution page
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Solution: From 1 3 2 2 3 \u00d7 + \u00d7 + \u00d7 ( ) ( ) ( ) : i ii iii we get, \u0394 = \u2212 ( ) + ( ) + \u2212 ( ) = \u2212 H kJ Required 1 3 46 4 2 9 0 2 3 41 24 8 . . .
Answer: C
\nOriginal PDF solution page
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Solution: For the reaction, 1 2 1 2 2 2 H g I s HI g ( ) + ( ) \u2192 ( ) ; \u0394 = \u2212 ( ) \u2212 \u2212 ( ) + \u2212 ( ) \u2212 \u2212 ( ) + \u2212 ( H Required 1 2 44 20 1 2 52 42 17 31 19 21 13 74 . . . . . ) ) \u2212 \u2212 ( ) = 13 67 5 94 . . kcal
Answer: A
\nOriginal PDF solution page
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Solution: The required thermochemical equation is I s O g I O s 5 2 2 2 5 2 ( ) + ( ) \u2192 ( ) From 2 \u00d7 (ii) + 6 \u00d7 (v) + 5 \u00d7 (vii) \u2013 (i) \u2013 6 \u00d7 (iii) \u2013 6 \u00d7 (iv) \u2013 (vi) \u2013 10 \u00d7 (viii) \u2013 10 \u00d7 (ix), we get, \u0394 = \u2212 ( ) + \u2212 ( ) + \u2212 ( ) \u2212 ( ) \u2212 \u2212 ( ) \u2212 \u2212 ( ) \u2212 \u2212 H Required 2 322 6 100 5 255 4 0 6 44 6 57 22 . 4 4 10 92 10 75 169 ( ) \u2212 \u2212 ( ) \u2212 \u2212 ( ) = \u2212 kJ
Answer: B
\nOriginal PDF solution page
Open page 275 in PDFExtracted text
Solution: Given thermochemical equations are (i) H g H g H kJ 2 ( ) \u2192 ( ) = 2 218 ; \u0394 (ii) Cl g 2Cl g H kJ 2 ( ) \u2192 ( ) = ; \u0394 124 (iii) 1 2 3 2 46 N g H g NH g H kJ 2 2 3 ( ) + ( ) \u2192 ( ) = \u2212 ; \u0394 (iv) 1 2 2 1 2 314 N g H g Cl g NH Cl s H kJ 2 2 2 4 ( ) + ( ) + ( ) \u2192 ( ) = \u2212 ; \u0394 (v) H g H g e H kJ ( ) \u2192 ( ) + = + \u2212 ; \u0394 1310 (vi) Cl g e Cl g H kJ ( ) + \u2192 ( ) = \u2212 \u2212 \u2212 ; \u0394 348 (vii) NH Cl s NH g Cl g H kJ 4 ( ) \u2192 ( ) + ( ) = + \u2212 4 683 ; \u0394 Required thermochemical equations are NH g H g NH g H 3 ( ) + ( ) \u2192 ( ) = + + 4 ; ? \u0394 From ( ) ( ) ( ) ( ) ( ) ( ) ( ) vii i v iii ii i v vi + \u2212 \u2212 \u2212 \u2212 \u2212 1 2 1 2 \u0394 = ( ) + \u2212 ( ) \u2212 \u2212 ( ) \u2212 \u00d7 ( ) \u2212 \u00d7 ( ) \u2212 ( ) \u2212 \u2212 H required 683 314 46 1 2 124 1 2 218 1310 348 8 718 ( ) = \u2212 kJ/mol\n5.39 Thermochemistry HINTS AND EXPLANATIONS
Answer: A
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: In such polymerization, one sigma bond is formed on cleavage of one pi bond. \u0394 H B.E. B.E. required C C bond C C bond = ( ) \u2212 ( ) = \u2212 ( ) \u2212 ( ) = \u2212 \u2212 \u2212 \u03c0 \u03c3 590 331 331 7 72 kJ/mole
Answer: D
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: Required thermochemical equation is C S 2H g O g CH OH l 2 2 3 ( ) + ( ) + ( ) \u2192 ( ) 1 2 \u0394 H kJ = + \u00d7 + [ ] \u2212 \u00d7 + + [ ] \u2212 = \u2212 715 4 218 249 3 415 356 463 38 266
Answer: B
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: 3 3 C s H g C H g H 53kJ 2 3 6 exp ( ) + ( ) \u2192 ( ) = ; \u0394 \u0394 H 3 715 6 218 3 356 6 408 63kJ theo = \u00d7 + \u00d7 [ ] \u2212 \u00d7 + \u00d7 [ ] = \u2212 \u2234 Strain energy H H kJ theo = \u2212 = \u0394 \u0394 exp 116
Answer: B
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: 2C g 6H g C H g 2 6 ( ) + ( ) \u2192 ( ) \u0394 H B.E B.E kJ C C C C = \u2212 = + [ ] \u2212 + \u00d7 [ ] \u21d2 = \u2212 \u2212 2839 0 0 6 412 367 and, 2C g 4H g C H g 2 4 ( ) + ( ) \u2192 ( ) \u0394 H B.E B.E. kJ C C C C = \u2212 = + [ ] \u2212 + \u00d7 [ ] \u21d2 = = = 2275 0 0 4 412 627 Now, 6C g 6H g C H g 6 6 ( ) + ( ) \u2192 ( ) \u0394 H R.E. = \u2212 = + [ ] \u2212 \u00d7 + \u00d7 + \u00d7 [ ] \u2212 5506 0 0 3 367 3 627 6 412 \u2234 R.E. kJ/mol = 52
Answer: A
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: C H S C H g S g C H S S C H g 2 5 2 5 2 5 2 5 \u2212 \u2212 ( ) + ( )\u2192 \u2212 \u2212 \u2212 ( ) \u0394 H kJ = \u2212 ( ) \u2212 \u2212 ( ) + \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 202 143 222 276 B.E. kJ/mol s s \u2212 = 276
Answer: B
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: CH g CH g H g 4 3 ( ) \u2192 ( ) + ( ) 103 103 2 18 33 5 = + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 \u2212 [ ] \u21d2 = ( ) ( ) \u0394 \u0394 f CH g f CH g H H kcal/mol 3 3 .
Answer: D
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: \u0394 H required = \u00d7 + \u00d7 + + \u00d7 [ ] \u2212 \u00d7 + + + + + 6 414 2 348 580 2 610 3 414 348 580 354 462 11 18 2 580 140 2 462 + \u00d7 + + \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u2212348 \u039a J
Answer: D
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: \u0394 H kJ = \u2212 = \u2212 50 70 20
Answer: A
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: \u0394 H B.E. B.E. bond in C C bond inC C = ( ) \u2212 ( ) = \u2212 ( ) \u2212 ( ) = \u2212 = \u2212 \u03c0 \u03c3 835 610 348 123 k kJ
Answer: B
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: n HCHO g HCHO 2 ( ) \u2192 ( ) \u0394 H n n = \u2212 = \u00d7 \u2212 ( ) \u2212 \u2212 ( ) \u21d2 = 72 134 732 6 \u2234 Molecular formula HCHO C H O 6 6 12 6 = ( ) =
Answer: B
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: There is 2.5 B-B bond per B atom. Hence, \u0394 H = \u20132.5 \u00d7 300 = \u2013750 kJ/mole of Boron.
Answer: C
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: Let the enthalpy of combustion of gauche form be \u2013 x kcal/mol. Now, 690 0 7 2 0 2 0 06 3 0 04 5 5 = \u00d7 \u2212 ( ) + \u00d7 + \u00d7 + ( ) + \u00d7 + ( ) . . . . . x x x x \u2234 x = 691
Answer: A
\nOriginal PDF solution page
Open page 276 in PDFExtracted text
Solution: M s X g MX s H 1.5 H eg g ( ) + ( ) \u2192 ( ) \u0394 = \u00d7 \u0394 ( ) 2 2 ; x From Born\u2013Hafer Cycle, we get: \u0394 = \u0394 + \u0394 + \u0394 + \u0394 + \u00d7 \u0394 + \u0394 ( ) ( ) ( ) ( ) ( ) H H H H H H sub M s i M g i M g Bond X g eg X g lat 1 2 2 2 l lice MX s H 2 ( ) or, 1 5 96 1 2 2 8 0 8 1 2 . . . . . \u00d7 \u2212 ( ) = \u0394 + \u00d7 \u0394 + \u00d7 \u0394 + \u00d7 \u00d7 \u0394 ( ) ( ) ( ) sub M s sub M s sub M s s H H H u ub M s sub M s H H ( ) ( ) ( ) + \u2212 ( ) + \u00d7 \u2212 \u00d7 \u00d7 \u0394 \u23a1 \u23a3 \u23a4 \u23a6 2 96 5 0 8 1 2 . . \u0394 = ( ) sub M s H kcal/mol 41 38 .\n5.40 Chapter 5 HINTS AND EXPLANATIONS
Answer: B, C, D
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: Combustion is exothermic. Decomposition or elimination are endothermic. Graphite is more stable form.
Answer: A, B, C
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: Conversion of liquid into gas is endothermic.
Answer: A, D
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: \u0394 \u00b0 = f H 0 for elements in their reference state.
Answer: A
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: Endothermic compounds have +ve \u0394 \u00b0 f H .
Answer: B
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: For \u0394 = \u0394 \u0394 = H E, n g 0
Answer: A
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: One mole of the substance should burn completely.
Answer: A, B
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: \u0394 = + ( ) f NO g H ve
Answer: A, B, D
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: Heat released in reaction = Heat gained by calorimeter system = \u00d7 = 1 5 1 4 2 1 . . . kJ n H SO eq 2 4 100 0 5 1000 0 05 ( ) = \u00d7 = . . n NH OH Limiting reagent eq 4 200 0 2 1000 0 04 ( ) = \u00d7 = . . ( ) \u0394 neut NH OH H By strong acid kJ/eq kJ/mole 4 ( ) . . . . = \u2212 = \u2212 = \u2212 2 1 0 04 52 5 52 5 \u0394 = \u2212 ( ) \u2212 \u2212 ( ) = diss NH OH H kJ/mol 4 52 5 57 4 5 . . \u0394 = \u2212 ( ) \u2212 = diss CH COOH H kJ/mol 3 57 48 1 4 5 4 4 . . .
Answer: A, D
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: (a) \u0394 = + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 = r H kJ 436 1 2 495 242 925 5 ( ) . (b) \u0394 = \u00d7 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 = r H kJ 1 2 436 1 2 495 42 423 5 ( ) . (c) \u0394 = \u00d7 = f H(g) H kJ mol 1 2 436 218 / (d) \u0394 = f OH(g) H kJ/mol 42
Answer: B, D
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: Resonance occurs in 1, 3-Butadiene and N 2 O.
Answer: A, B, C
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: Resonance occurs in product but not in reactant.
Answer: A, B
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: (a) \u0394 = \u00d7 \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 r H kJ 2 1263 2238 285 3 (b) \u2212 = \u0394 \u2212 \u00d7 \u0394 3KJ H H C -maltose C glucose \u03b1 2
Answer: A, B, C
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: (a) n C s ( ) = \u00d7 = 1 2 1000 12 100 . \u2234 Maximum obtainable heat = 100 \u00d7 94 = 9400 cal (b) Heat released = \u00d7 + \u00d7 = 100 68 100 68 13600 cal (c) Heat released = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 13600 100 1200 30 5440 cal
Answer: A, D
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: C H COOH s O g CO g H O l 2 6 5 2 2 15 2 7 3 ( ) + ( ) \u2192 ( ) + ( ) \u0394 = \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 [ ] = \u2212 H kJ/mol 7 393 3 286 408 3201 and \u0394 = \u0394 \u2212 \u0394 = \u2212 ( ) \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 = \u2212 U H n RT kJ/ g . . . 3201 7 15 2 8 314 1000 300 3199 75 m mol
Answer: C
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: \u0394 H kcal = = \u00d7 \u2212 ( ) = \u2212 q 3 35 105 \u0394 \u0394 \u0394 U H n RT kcal g = \u2212 = \u2212 ( ) \u2212 \u2212 ( ) \u00d7 \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u00d7 = \u2212 . . 35 2 3 2 1000 300 3 103 2 and w u q = \u2212 = \u2212 ( ) \u2212 ( ) = 103 2 105 1 8 . . kcal
Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: From \u2212 \u00d7 \u2212 \u00d7 + \u00d7 \u2212 \u00d7 3 4 1 4 1 4 9 4 a b c d , \u0394 = \u2212 \u00d7 \u2212 \u2212 \u00d7 + \u00d7 \u2212 = \u2212 H kcal required 3 4 76 1 4 240 1 4 36 9 4 68 147 ( ) ( ) ( ) ( )
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: From 3 4 1 4 1 4 1 4 \u00d7 + \u00d7 \u2212 \u00d7 + \u00d7 a bc d , \u0394 = \u00d7 \u2212 + \u00d7 \u2212 \u00d7 + = H kcal/mol required 3 4 76 1 4 240 1 4 36 1 4 68 11 ( ) ( ) ( ) ( )
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 277 in PDFExtracted text
Solution: Given data based\n5.41 Thermochemistry HINTS AND EXPLANATIONS Comprehension II
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: \u0394 = \u2212 \u00d7 + \u22c5 = \u2212 \u22c5 H kJ 75 5 5 1 8 55 15
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: \u0394 = \u2212 \u00d7 + \u22c5 = \u2212 \u22c5 H kJ 75 10 10 1 8 63 56
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: \u0394 = \u2212 + \u22c5 = \u2212 + \u22c5 \u221e = \u2212 H n kJ 75 1 1 8 75 1 1 8 75
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: \u0394 = \u2212 \u2212 \u2212 = \u2212 H kJ ( . ) ( . ) . 63 56 55 15 8 41
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: \u0394 = \u2212 \u2212 \u2212 = \u2212 H kJ ( ) ( . ) . 75 63 56 11 44 Comprehension III
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/mol H 1 2 483 636 2 241 818 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/mol H 2 2 868 2 3 289 4 \u0394 = \u2212 H kJ/mol H 3 2 347 33 .
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm O 1 2 483 636 32 15 11 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm O 2 3 868 2 48 18 09 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm H O 3 2 2 347 33 34 10 22
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm reactant 1 483 636 36 13 43 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm reactant 2 868 2 54 16 08 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm reactant 3 347 33 36 9 65 Comprehension IV
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: Heat released = 0.25 \u00d7 320 = 80 cal \u2234 Molar enthalpy of solution = \u2212 \u22c5 \u00d7 = \u2212 80 0 98 98 8000 cal
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: Heat released = 0.80 \u00d7 320 = 256 cal \u2234 \u0394 \u2212 \u22c5 \u00d7 = \u2212 r H = cal 256 0 49 98 51200 Comprehension V
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 278 in PDFExtracted text
Solution: 2 3 2 1 2 2 2 3 C(S)+ H g N g CH CN(g); ( ) ( ) + \u2192 \u0394 H B.E. B.E. C C C N = = \u00d7 + \u00d7 + \u00d7 \u2212 \u00d7 + + \u2192 \u2212 \u2261 88 2 719 3 2 435 1 2 948 3 414 1 [ ] [ ] 3 2 3 8 C(s) 4H g C H g + \u2192 ( ) ( ) \u0394 = \u2212 = \u00d7 + \u00d7 \u2212 \u00d7 + \u00d7 \u2192 \u2212 H B.E. C C 85 3 719 4 435 2 8 414 2 [ ] [ ] From (1) and (2), we get: B.E. kJ/mol and B.E. kJ/mol C C C N \u2212 \u2261 = = 335 899 5 . Now, CH CN(g) 2H g CH CH NH (g), 3 2 3 2 2 + \u2192 ( ) \u0394 = \u00d7 + + + \u00d7 \u2212 \u00d7 + + + \u00d7 H [ . ] [ ] 3 414 335 899 5 2 435 5 414 335 378 2 426 = \u2212 288 5 . kJ/mol\n5.42 Chapter 5 HINTS AND EXPLANATIONS Comprehension VI
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: C < C p,m,N (g) p,m,H O(g) 2 2
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: H g O g H O g 2 2 2 1 2 ( ) ( ) ( ); + \u2192 \u0394 = \u2212 \u22c5 \u0394 = \u2212 \u22c5 H kcal U kcal 55 85 56 0 Let x mole H 2 be burnt. x 2 mol O (g) 2 is needed and hence, x 2 4 2 \u00d7 = \u00d7 mol N 2 is also present. Now, Heat released from reaction = Heat gained by H O(g) 2 and N g 2 ( ) 56.0 \u00d7 10 3 = x \u00d7 6.2 \u00d7 ( T 2 \u2013 300) + 2 x \u00d7 4.9 \u00d7 ( T 2 \u2013 300) \u2234 T 2 = 3800 K
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: p n T p n T x x x p x x 1 1 1 2 2 2 2 1 2 2 300 2 3800 = \u21d2 + + \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = + \u00d7 ( ) \u2234 p 2 = 10.86 atm
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: q = 0, w = 0 \u21d2 \u2206 E = 0 Comprehension VII
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: C H l O g CO g H O(g) 18 8 2 2 2 25 2 8 9 ( ) ( ) ( ) + \u2192 + \u2206 c H = [8 \u00d7 (\u221294) + 9 \u00d7 (\u221258)] \u2212 [\u221274] = \u22121200 kcal/mol
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: \u2206 H required = [8 \u00d7 (\u221226.5) + 9 \u00d7 (\u221258)] \u2212 [\u221274] = \u2212660 kcal/mol
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: Let x mole of C 8 H 18 be converted into CO 2 . As temperature is increased, some heat is absorbed by product gases. Now, [ ( ) ] [ ( ) x x x x \u00d7 + \u22c5 \u2212 \u00d7 \u2212 \u00d7 \u00d7 \u00d7 + \u22c5 \u2212 \u00d7 \u22c5 \u00d7 + \u22c5 \u00d7 1200 0 1 660 1 1000 8 8 500 8 0 1 7 0 500 0 9 6 6 0 500 87 3 \u22c5 \u00d7 = \u22c5 ] \u2234 x = 0.05 Moles of CO 2 formed = 0.05 \u00d7 8 = 0.4
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: Moles of H 2 O formed = 9 x + (0.1\u2212 x ) \u00d7 9 = 0.9
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: w p v v p n RT p n RT p R n T n T = \u2212 \u2212 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 ( ) ( ) 2 1 2 2 1 1 2 2 1 1 = \u2212 \u00d7 \u22c5 \u00d7 + \u22c5 \u2212 \u22c5 \u00d7 + \u22c5 \u00d7 \u2212 \u22c5 \u00d7 + \u22c5 \u2212 \u22c5 \u00d7 2 0 05 8 0 1 0 05 8 0 9 800 0 05 25 2 0 1 0 05 17 2 [{ ( ) } ( ) { { } \u00d7 \u2212 300 2090 ] = cal Comprehension VIII
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: CO 2H CH OH rd mol + \u23af \u2192 \u23af\u23af 2 2 3 3 1000 = \u00d7 = 3 2 1000 1500 mol In reformer, CO and H 2 is forming in 1 : 3 ratio.
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: CO = 1500 \u22121000 = 500 mole H 2 = 4500 \u2212 2000 = 2500 mole
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 279 in PDFExtracted text
Solution: Heat produced in 1 min = 1000 \u00d7 100 R \u00d7 60 = 1.2 \u00d7 10 7 cal\n5.43 Thermochemistry HINTS AND EXPLANATIONS Comprehension IX
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: Let x mole C be converted into CO. Hence, x \u00d7 26 + (1 \u2013 x ) \u00d7 94 = 53.2 \u21d2 x = 0.6 Hence, moles of C formed = 0.6
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: O consumed +(1 ) ]32 gm 2 2 22 4 = \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u22c5 x x Comprehension X Given thermochemical equations are (a) H 2 S (g) \u2192 H (g) + H S (g); \u2206 H = 376.0 kcal (b) H 2 (g) + S(s) \u2192 H 2 S (g); \u2206 H = \u221220.0 kcal (c) S (s) \u2192 S (g); \u2206 H = 277.0 kcal (d) H 2 (g) \u2192 2H (g); \u2206 H = 436.0 kcal
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: 1 2 2 H S(s) HS(g) ( ) g + \u2192 From a b d we get: + \u2212 \u00d7 1 2 , \u0394 = + \u2212 \u2212 \u00d7 = H kJ mol required 376 20 1 2 436 138 ( ) /
Comprehension 10 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: HS(g) \u2192 H(g) + S(g) From (d) \u2212 (a) \u2212 (b) + (c) \u2206 H required = 436 \u2212 376 \u2212 (\u221220) + 277 = 357 kJ/mol Comprehension XI
Comprehension 11 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: \u0394 \u00b0 = \u0394 \u00b0 + \u0394 \u22c5 = + \u00d7 \u00d7 = H E n RT kcal g 2 1 2 2 1000 298 3 292 . . Now, \u0394 \u00b0 = \u0394 \u00b0 \u2212 \u22c5 \u0394 \u00b0 = \u2212 \u00d7 = \u2212 G H T S kcal 3 292 298 1000 20 2 668 . .
Comprehension 11 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: Spontaneous as \u2206 G\u00b0 = \u2212ve
Comprehension 12 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
"
+ }
+ },
+ {
+ "question_id": "thermochemistry-chem-sec-3-34-82",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": 1,
+ "subject": "chemistry",
+ "chapter": "thermochemistry",
+ "chapterTitle": "Thermochemistry",
+ "type": "mcq",
+ "rawChapterType": "MCQ",
+ "originalNumber": 82,
+ "displayNumber": 34,
+ "image": "./assets/chapter-packs/thermochemistry/Chemistry Section 1__--__82__--__1.png",
+ "solutionImage": null,
+ "question": {
+ "content": "Comprehension 12 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
"
+ }
+ }
+ ]
+ },
+ {
+ "title": "Chem Sec 4",
+ "originalName": "Section D - Assertion Reason",
+ "questions": [
+ {
+ "question_id": "thermochemistry-chem-sec-4-1-83",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": 1,
+ "subject": "chemistry",
+ "chapter": "thermochemistry",
+ "chapterTitle": "Thermochemistry",
+ "type": "mcq",
+ "rawChapterType": "MCQ",
+ "originalNumber": 83,
+ "displayNumber": 1,
+ "image": "./assets/chapter-packs/thermochemistry/Chemistry Section 1__--__83__--__1.png",
+ "solutionImage": null,
+ "question": {
+ "content": "",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Answer: D
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: H 2 SO 4 is dibasic but HCl is monobasic.
Answer: D
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: Information based
Answer: D
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: Heat liberated will be four times but as quantity is also four times, the change in the temperature will be same.
Answer: C
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: Solubility is exothermic but all gases are not highly soluble in all liquid.
Answer: C
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: If | \u2206 Hydration H| < | \u2206 lattice H|, the salt dissolves partially and the extent depends on the difference in two values.
Answer: A
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: H HDiamond + O2 Hgraphite + O2 HCO2
Answer: A
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: H H1 \u2013 Extent + O2 H2 \u2013 Extent + O2 HCO2 + H2O
Answer: C
\nOriginal PDF solution page
Open page 280 in PDFExtracted text
Solution: Theory based\n5.44 Chapter 5 HINTS AND EXPLANATIONS
Answer: A \u2192 R; B \u2192 P; C \u2192 Q; D \u2192 P, R, S
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: C \u2192 CO, \u2206 H \u2260 \u2206 H combustion
Answer: A \u2192 Q; B \u2192 S; C \u2192 P; D \u2192 R
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: (A) \u2206 n g = 0 (B) \u2206 n g = \u22121 (C) \u2206 n g = 1 (D) \u2206 n g = \u2212 2
Answer: A \u2192 P, B \u2192 Q; C \u2192 R; D \u2192 S
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: (A) \u2206 H = (\u221257.3) + 15 = \u221242.3 kJ (B) \u2206 H = \u221242.3 \u2212 70.7 + 20 = \u221293.0 kJ (C) \u2206 H = \u221270.7 + 15 = \u221255.7 kJ (D) \u2206 H = 0
Answer: A \u2192 Q; B \u2192 R; C \u2192 P; D \u2192 S
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: (A) Mg (s) + Cl2 (g) \u2013(1110 + 790) (2360 \u2013 1110) Mg 2+ (aq) + 2Cl \u2013 (aq) Mg 2+ (g) + 2Cl \u2013 (aq) + 1110 Mg 2f (aq); +Cl2 (g) or = 1110 \u2013 (1110 + 790) + (2360 \u2013 1110) = 460 kJ/mol (B) 1 2 1 2 1110 2360 1 2 1 2 652 2 2 Cl g Cl ag H Mg g Mg g 2 ( ) ( ); ( ) ( ) ( ) \u2192 \u0394 = \u2212 + + = \u2212 \u2212 + + K KJ/mol (C) Mg 2+ (g) +2Cl \u2212 (aq) \u2192 Mg 2+ (aq) +2Cl \u2212 (aq); \u2206 H = \u2212790 \u2212 1110 = \u22121900 kJ (D) Mg 2+ (g) + 2Cl \u2212 (g) \u2192 MgCl 2 (s); \u2206 H = \u2212 640 \u2212 1870 = \u2212 2510 kJ
Answer: A \u2192 P, Q; B \u2192 P, R; C \u2192 S
\nOriginal PDF solution page
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Solution: Defi nition based
Answer: A \u2192 Q; B \u2192 R; C \u2192 P; D \u2192 S
\nOriginal PDF solution page
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Solution: 3O 2 (g) \u2192 2O 2 (g)
Answer: A \u2192 S; B \u2192 Q; C \u2192 P; D \u2192 R
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Answer: A \u2192 P; B \u2192 Q, T; C \u2192 Q, S; D \u2192 R, S
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: \u2206 n g = 0 \u21d2 \u2206 H = \u2206 U \u2206 n g = + ve \u21d2 \u2206 H > \u2206 U \u21d2 If \u2206 H = \u2212ve, then | \u2206 H| < | \u2206 U| \u21d2 If \u2206 H = +ve, then | \u2206 H| > | \u2206 U| \u2206 n g = \u2212 ve \u21d2 \u2206 H < \u2206 U \u21d2 If \u2206 H = \u2212ve, then | \u2206 H| > | \u2206 U| \u21d2 If \u2206 H = +ve, then | \u2206 H| < | \u2206 U|
Answer: A \u2192 P; B \u2192 P, Q; C \u2192 R, S; D \u2192 R
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: Defi nition based (10) (A) \u0394 = \u0394 + \u2212 \u00d7 \u2212 = + \u2212 \u00d7 \u00d7 H H c c T T p,A(g p,A(l 400 300 2 1 25 20 40 1 1000 40 [ ] [ ] ( ) ( ) ) 0 0 300 23 \u2212 = + ) kJ/mol (B) \u0394 = \u0394 + \u2212 \u00d7 \u2212 = + \u2212 \u00d7 \u00d7 H H c c T T p,A (g p,A (l 300 400 2 1 3 3 50 30 50 1 1000 [ ] [ ] ( ) ( ) ) 3 300 400 52 \u2212 = + ) kJ/mol (C) \u2206 H 300 = 3 \u00d7 25 \u2212 100 \u2212 52 = \u2212 77 kJ/mol (D) \u0394 = \u0394 + \u2212 \u00d7 \u00d7 \u2212 = \u2212 + \u2212 \u00d7 \u00d7 H H c c T T p,A (l p,A (l 400 300 2 1 3 3 3 77 50 3 40 [ ] [ ] ( ) ( ) ) ) 1 1 1000 400 300 84 \u00d7 \u2212 = \u2212 ( ) kcal/mol
Answer: A \u2192 Q; B \u2192 R; C \u2192 S; D \u2192 P
"
+ }
+ }
+ ]
+ },
+ {
+ "title": "Chem Sec 6",
+ "originalName": "Section F - Integer",
+ "questions": [
+ {
+ "question_id": "thermochemistry-chem-sec-6-1-103",
+ "marks": 4.0,
+ "negMarks": 1.0,
+ "partialMarks": null,
+ "subject": "chemistry",
+ "chapter": "thermochemistry",
+ "chapterTitle": "Thermochemistry",
+ "type": "sa",
+ "rawChapterType": "NAT",
+ "originalNumber": 103,
+ "displayNumber": 1,
+ "image": "./assets/chapter-packs/thermochemistry/Chemistry Section 1__--__103__--__1.png",
+ "solutionImage": null,
+ "question": {
+ "content": "",
+ "options": [],
+ "correct_options": [],
+ "answer": "3",
+ "explanation": "Answer: 3
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: For HCl : 13.7 \u00d7 0.05 = c \u00d7 411 (1) For HCOOH : q \u00d7 0.05 = c \u00d7 321 (2) From (2) \u00f7 (1) \u21d2 q = 10.7 kcal \u2234 Enthalpy of ionisation of HCOOH = 13.7 \u2212 10.7 = 3.0 kcal/mol
Answer: 1
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: (C 6 H 10 O 5 ) x + 6 x O 2 (g) \u2192 6 x CO 2 (g) + 5 x H 2 O(l) \u2206 H = \u2212 4.6 \u00d7 162 x = [6x( \u221294.2) + 5 x (\u221268.4)] \u2212 \u0394 f C H O H 6 10 5 ( ) x + \u23a1 \u23a3 \u23a4 \u23a6 0 \u2234 \u0394 f C H O H 6 10 5 ( ) x = \u2212162 kcal/mol = \u22121kcal/gm
Answer: 5
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: 1 800 3120 800 3120 \u00d7 = \u00d7 \u21d2 = a a L/hr Butane C H O CO H O 2 4 10 2 2 13 2 4 5 + \u2192 + \u2234 Rate of Oxygen Supply L/hr = \u00d7 \u00d7 = 800 3120 13 2 3 5
Answer: 2
\nOriginal PDF solution page
Open page 281 in PDFExtracted text
Solution: \u0394 = \u00d7 \u2212 = H kcal/mol required 100 75 13 7 12 2 2 ( . . )\n5.45 Thermochemistry HINTS AND EXPLANATIONS
Answer: 5
\nOriginal PDF solution page
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Solution: Total moles of gases = \u00d7 \u00d7 = 1 192 1 642 0 0821 298 0 08 . . . . Now, n n CH CH 4 4 210 10 1260 0 667 0 004 3 \u00d7 \u00d7 = \u00d7 \u21d2 = . . \u2234 Volume per cent of CH 4 = \u22c5 \u22c5 \u00d7 = 0 004 0 08 100 5%
Answer: 8
\nOriginal PDF solution page
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Solution: Heat released by 6 3 64000 \u22c5 mole haemoglobin = 25 4.2 J \u00d7 \u00d7 = 0 03 3 15 . . \u2234 Heat released per mole haemoglobin = \u22c5 \u00d7 \u22c5 = 3 15 64000 6 3 32000 J \u2234 Heat released per mole O 2 = = 32000 4 8000 J
Answer: 6
\nOriginal PDF solution page
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Solution: Heat released = \u00d7 \u00d7 \u00d7 \u2212 = 300 1 0 1 0 26 25 300 . . ( ) cal Now, n HA = \u00d7 \u22c5 = \u22c5 200 0 4 1000 0 08 n NaOH = \u00d7 \u22c5 = \u22c5 100 0 5 1000 0 05 Hence, NaOH is a limiting reagent. \u2234 \u0394 = \u2212 \u22c5 \u00d7 = \u2212 neut H cal/mol 300 0 05 1 6000
Answer: 5
\nOriginal PDF solution page
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Solution: There is 3 H-bond per NH 3 molecule because for each bond two NH 3 molecules are required. \u2234 Strength of H-bond = \u2212 = 30 4 15 4 3 5 0 . . . kcal/mol
Answer: 6
\nOriginal PDF solution page
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Solution: C(s)+ O g CO(g); H kcal/mol 2 1 1 2 7 5 3 12 30 ( ) \u2192 \u0394 = \u2212 \u22c5 \u00d7 = \u2212 C(s)+O g CO (g); H kcal/mol 2 2 ( ) \u2192 \u0394 = \u2212 \u00d7 = \u2212 2 32 4 12 96 Now, CO (g) CO(g)+ O g H H H kcal/mol 2 1 2 2 1 2 66 \u2192 \u0394 = \u0394 \u2212 \u0394 = + ( ); For 4 gm CO H kcal 2 \u22c5 \u0394 = \u00d7 = 0 66 44 4 6 ,
Answer: 3
\nOriginal PDF solution page
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Solution: ( ) 1 6900 3 4 \u2212 \u00d7 + \u00d7 = \u21d2 = a a a 2900 3900 \u2234 n n eq(HA) eq(HB) : ( ) : : : = \u2212 = = 1 1 4 3 4 1 3 a a Four-digit Integer Type
Answer: 0043
\nOriginal PDF solution page
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Solution: C 2 H 6 + H 2 \u2192 2 CH 4 ; \u2206 H = \u221265.2 kJ C 3 H 8 + 2H 2 \u2192 3 CH 4 ; \u2206 H = \u221287.4 kJ Hence, for CH 4 (g) + C 3 H 8 (g) \u2192 2 C 2 H 6 (g); \u2206 H = (\u221287.4) \u22122 \u00d7 (\u221265.2) = + 43 kJ
Answer: 0400
\nOriginal PDF solution page
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Solution: Moles of O 2 consumed = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u23a7 \u23a8 \u23a9 \u23ab \u23ac \u23ad \u00d7 \u22c5 \u00d7 = 164 2 1000 20 10 100 20 60 1 0 0821 310 24 31 . C H O O CO HO H = kJ 6 2 2 2 12 6 6 6 6 3100 + \u2192 + \u0394 \u2212 ; \u2234 Heat produced in body per hr = \u00d7 = 3100 6 24 31 400 kJ
Answer: 0030
\nOriginal PDF solution page
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Solution: Number of glycogen units oxidized per day = \u00d7 \u00d7 \u00d7 \u00d7 = 150 60 60 24 432 10 30 3
Answer: 0085
\nOriginal PDF solution page
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Solution: Moles of C = = 15 12 1 25 . Moles of O 2 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u22c5 \u00d7 = 20 19 100 8 21 0 0821 380 1 . 1 25 0 5 2 2 . . C + O CO +0.75 CO \u2192 \u2234 Heat produced = \u00d7 + \u00d7 = 0 5 26 0 75 96 85 . . kcal\n5.46 Chapter 5 HINTS AND EXPLANATIONS
Answer: 2400
\nOriginal PDF solution page
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Solution: C 2 H 5 OH(l) + O 2 (g) \u2192 CH 3 COOH(g) + H 2 O(l) \u2206 H = [(\u2212118)+(\u221268)] \u2212 [(\u221266)+0] = \u2212120 kcal Hence, rate of heat removal = \u00d7 \u00d7 \u00d7 = 120 46 2 3 10 40 100 2400 3 . kcal/mol
Answer: 0216
\nOriginal PDF solution page
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Solution: C 6 H 12 O 6 (s) + 6 O 2 (g) \u2192 6 CO 2 (g) + 6 H 2 O(l) \u2206 H = [6 \u00d7 (\u2212395) + 6 \u00d7 (\u2212285)] \u2212 [(\u22121280) + 0] = \u22122800 kJ Moles of CO 2 released per astronaut = \u00d7 = 6 2800 2100 4 5 . \u2234 Mass of LiOH required = 4.5 \u00d7 2 \u00d7 24 = 216 gm
Answer: 0500
\nOriginal PDF solution page
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Solution: 16 1 322 100 10 500 3 3 . . \u00d7 \u00d7 = \u00d7 \u21d2 = = V 10000 V 0 5m L
Answer: 0120
\nOriginal PDF solution page
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Solution: Total heat absorbed = \u00d7 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 8 9 45 1 9 72 2 5 120 . KJ
Answer: 0075
\nOriginal PDF solution page
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Solution: For banana: q = c \u00d7 3.0 (1) For benzoic acid: 800 122 0 305 0 \u00d7 = \u00d7 . . c 4 (2) From (1) and (2), q = 1.5 kacl for 2.5 gm banana \u2234 Heat obtained per banana = \u22c5 \u22c5 \u00d7 = 1 5 2 5 125 75 kcal
Answer: 0241
\nOriginal PDF solution page
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Solution: H O H O(l); H kJ 2 2 2 298 1 2 286 ( ) ( ) g g + \u2192 \u0394 = \u2212 H O(l) H O(g); H kJ 2 2 \u2192 \u0394 = \u22c5 398 40 8 \u0394 = \u0394 + \u0394 \u22c5 \u0394 = \u2212 \u00d7 \u2212 = H H C T 40.8+ kJ 398 p 298 33 4 75 4 1000 298 398 45 . . ( ) \u2234 H O(g) H O H 2 2 \u2192 + \u0394 = \u2212 \u2212 2 298 1 2 45 286 ( ) ( ); [ ] g g = 241 kJ
Answer: 0700
\nOriginal PDF solution page
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Solution: \u0394 \u2212 \u0394 = \u0394 \u22c5 \u222b H H C dT p T T 1 2 1 2 or, 0 T dT T T \u2212 \u2212 = \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u222b ( ) ( ) ( ) 4000 2 10 2 10 2 300 2 300 2 2 2 \u2234 T K = 700
Answer: 0120
\nOriginal PDF solution page
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Solution: 3C(s) + 3H 2 (g) \u2192 C 3 H 6 (g) \u2206 H theo = (3 \u00d7 715 + 6 \u00d7 218) \u2212 (3 \u00d7 356 + 6 \u00d7 408) = \u221263 kJ \u2206 H exp = [3 \u00d7 (\u2212393) +3 \u00d7 (\u2212285)] \u2212 [3 \u00d7 (\u2212697)] = 57 kJ \u2234 Strain energy = 57 \u2212 (\u221263) = 120 kJ/mol
Answer: 0060
\nOriginal PDF solution page
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Solution: KF.CH COOH s K g F CH COOH g H 734 kJ 3 3 ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 . ; CH COOH l CH COOH g H 20 kJ 3 3 ( ) \u2192 ( ) \u0394 = ; KF s K aq F aq H kJ ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 ; 35 K g K aq H kJ + + ( ) \u2192 ( ) \u0394 = \u2212 ; 325 F g F aq H kJ \u2212 \u2212 ( ) \u2192 ( ) \u0394 = \u2212 ; 389 KF s CH COOH l KF CH COOH s H kJ ( ) + ( ) \u2192 ( ) \u0394 = \u2212 3 1 3 25 ; Required: F g CH COOH g F CH COOH g \u2212 \u2212 ( ) + ( ) \u2192 ( ) 3 3 ; \u0394 = \u2212 ( ) + \u2212 \u2212 + \u2212 ( ) + \u2212 ( ) = \u2212 H kJ/mol 389 734 20 35 325 25 60
Answer: 0455
\nOriginal PDF solution page
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Solution: B s H BH g ( ) + ( ) \u2192 ( ) 3 2 2 3 g \u0394 = = + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 ( ) \u2212 H B E. B H 100 565 3 2 436 3 . \u2234 B E. kJ/mol B H . \u2212 = 373 2B(s) + 3H g B H g 2 ( ) ( ) \u2192 2 6 \u0394 = = \u00d7 + \u00d7 [ ] \u2212 \u00d7 + \u00d7 [ ] \u2212 H B E. 3c 2e 36 2 565 3 436 4 373 2 . \u2234 B E. kJ/mol 3c 2e . \u2212 = 455\n5.47 Thermochemistry HINTS AND EXPLANATIONS
Answer: 0292
\nOriginal PDF solution page
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Solution: XeF Xe F F F H kcal 4 + \u2192 + + + \u0394 = \u00d7 ( ) + + \u2212 ( ) + \u2212 ( ) = \u2212 2 4 34 279 85 38 292
Answer: 0021
\nOriginal PDF solution page
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Solution: (a) KF CH COOH s K ACOH F ACOH CH COOH l kJ . . 3 3 3 ( ) \u2192 ( ) + ( ) + ( ) = + \u2212 (b) KF s K ACOH F ACOH H 35 kJ ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 ; (c) F CH COOH g F g CH COOH g H 46 kJ . ; \u2212 \u2212 ( ) \u2192 ( ) + ( ) \u0394 = 3 3 (d) KF CH COOH s K g F CH COOH g H 734 kJ . . ; 3 3 ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 (e) KF s K g F g H 797 kJ ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 ; Required: CH COOH l CH COOH g 3 3 ( ) \u2192 ( ) From (c) (a)+(d) e b \u2212 \u2212 + ( ) ( ), we get: \u0394 = \u2212 \u2212 ( ) + \u2212 + = H 46 kJ/mol 3 734 797 35 21
Answer: 0085
\nOriginal PDF solution page
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Solution: (l) + 3H2 (g) (g) N NH \u2206 H = (\u221250) \u2212 [ \u2206 f H py(l) + 0] = (40 + 125) + [2 \u00d7 {(\u2212156) \u2212 (\u221237)} + {(\u221218) \u2212 44}] \u2234 \u2206 f H py(l) = 85 kJ/mol
Answer: 0272
\nOriginal PDF solution page
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Solution: 1 2 5 2 847 2 2 5 I F g IF g H kJ s ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; \u2212 = \u00d7 + ( ) + \u00d7 \u2212 \u00d7 \u2212 847 B E. I F 1 2 62 149 5 2 155 5 . B E. kJ/mol I F . \u2212 = 268 1 2 3 2 470 2 2 3 I s F g IF g H kJ ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; \u2212 = + ( ) + \u00d7 \u2212 \u00d7 + \u23a1 \u23a3 \u23a4 \u23a6 \u2212 ( ) 470 1 2 62 149 3 2 155 2 268 B E. I F eq . B E. kJ/mol I F eq . \u2212 ( ) = 272
Answer: 0142
\nOriginal PDF solution page
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Solution: H g O g H O l H kJ 2 2 2 1 2 286 ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; \u2212 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 \u2212 \u2212 \u2212 286 = B E. B E. H H O H . . 1 2 498 2 44 (1) H g O g H O l H kJ 2 2 2 2 188 ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; \u2212 + ( ) \u2212 \u00d7 + \u2212 \u2212 \u2212 \u2212 188 = B E. B E. B E. H H O H O O . ( . . ) 498 2 53 (2) From (1) (2), we get: B E. kJ/mol O O \u2212 = \u2212 . 142
Answer: 0120
\nOriginal PDF solution page
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Solution: (a) Ag + (aq) +Br - (aq) \u2192 AgBr(s); \u2206 H = \u221284.54 kJ (b) Ag(s) \u2192 Ag + (aq); \u2206 H = \u22128 x kJ (c) 1 2 9 Br l Br aq H kJ 2 ( ) \u2192 ( ) = \u2212 ; \u0394 x (d) Ag s Br l Ag Br s H kJ 2 ( ) + ( ) \u2192 ( ) = \u2212 1 2 99 54 ; . \u0394 As (a) + (b) + (c) = (d), we get: (\u221284.54) + (\u22128 x ) + 9 x = \u221299.54 \u21d2 x = \u221215 \u2234 \u2206 f H Ag + (aq) = \u22128 x = 120 kJ/mol
![\"Thermodynamics](\"./assets/chapter-packs/thermodynamics/Chemistry)
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
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Solution: U n f R T = \u00d7 \u00d7 2 For larger U , n , f , T , should be higher.
Answer: B
\nOriginal PDF solution page
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Solution: w P dv a V b dv a V V b V V V V V V = \u2212 \u22c5 = \u2212 + \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u2212 \u222b \u222b 1 2 1 2 2 1 2 1 ln ( )
Answer: A
\nOriginal PDF solution page
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Solution: q = m.s. \u0394 T \u21d2 10 \u00d7 10 6 = 80 \u00d7 (4.2 \u00d7 10 3 ) \u00d7 \u0394 T \u21d2 \u0394 T = 29.76 K = 29.76\u00b0 C
Answer: C
\nOriginal PDF solution page
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Solution: Heat lost by water = Heat gained by ice or, 500 75 6 18 20 9 6000 18 \u00d7 \u00d7 = \u00d7 \u00d7 . ( ) N \u21d2 N = 14
Answer: A
\nOriginal PDF solution page
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Solution: Let T 1 > T 2 . Now, heat lost by gas (1) = Heat gained by gas (2) or, n 1 \u22c5 C m \u22c5 ( T 1 \u2013 T f ) = n 2 \u22c5 C m \u22c5 ( T f \u2013 T 2 ) or, PV RT T T P V RT T T f f 1 1 1 1 2 2 2 2 \u22c5 \u2212 = \u2212 ( ) ( ) \u21d2 T T T PV P V PV T P V T f = + + 1 2 1 1 2 2 1 1 2 2 2 1 ( )
Answer: A
\nOriginal PDF solution page
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Solution: w = \u2013 P ext ( V 2 \u2013 V 1 ) = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f P nRT P nRT P 2 2 1 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f nRT P P nRT P P 1 1 1 2 1 1 1 = \u2212 \u00d7 nRT P 1 1\n4.37 Thermodynamics HINTS AND EXPLANATIONS Now, w total = w 1 + w 2 + \u2026 + w f = \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u2212 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f + + \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f nRT P nRT P nRT 1 1 1 1 2 \u001d = \u2212 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = = \u2212 \u2211 nRT P i i i P 1 1 1 1
Answer: B
\nOriginal PDF solution page
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Solution: w P dV K V dV K V V K V V V V V 1 1 2 1 0 0 2 1 0 2 2 1 2 1 4 0 0 1 2 = \u2212 \u22c5 = \u2212 \u22c5 = + \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u2212 \u222b \u222b = \u2212 = \u2212 = \u2212 2 4 2 0 5 0 0 2 0 0 0 0 0 P V V P V P V . w P dV K V dV K V V P V V V V V 2 1 2 0 0 2 0 0 2 0 7 0 0 1 2 = \u2212 \u22c5 = \u2212 \u22c5 = \u2212 = \u2212 \u00d7 \u222b \u222b ln .
Answer: A
\nOriginal PDF solution page
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Solution: As \u0394 T = 0, \u0394 U = 0
Answer: D
\nOriginal PDF solution page
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Solution: Area = P 2 \u00d7 \u0394 V \u21d2 P 2 \u00d7 4 = 49.26 L -atom Now, correct work, w = \u2212 \u22c5 = \u2212 \u22c5 nRT V V P V ln ln 2 1 2 2 4 2 = \u201349.26 \u00d7 0.693 = \u2013 34.137 L -atom
Answer: A
\nOriginal PDF solution page
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Solution: PV x = Constant \u21d2 = \u21d2 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u21d2 = PV P V P P V V x x x x x 1 1 2 2 1 2 2 1 8 4 3 2 Now, C C R x R R R m v m = + \u2212 = + \u2212 = \u2212 1 1 3 2 1 3 2 2
Answer: C
\nOriginal PDF solution page
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Solution: Dulong and Petit\u2019s law is applicable only for solid element. (Molar heat capacity \u2248 6.4 cal/K-mol = 26.8 J/K-mol).
Answer: C
\nOriginal PDF solution page
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Solution: PV P V 1 1 2 2 \u03b3 \u03b3 = \u21d2 P d P d 1 1 2 2 \u03b3 \u03b3 = \u21d2 P P d d 2 1 2 1 7 5 32 128 = \u239b \u239d \u239c \u239e \u23a0 \u239f = = \u03b3 ( ) /
Answer: C
\nOriginal PDF solution page
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Solution: U U T T rms rms , , 2 1 1 2 1 4 2 1 = = \u21d2 Now, T.V r \u2013 1 = Constant \u21d2 T T V V V V r 2 1 1 2 1 1 2 7 5 1 1 4 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 / \u2234 V 2 = 32 V 1
Answer: B
\nOriginal PDF solution page
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Solution: w w A B = \u00d7 2 But \u0394 U A = \u0394 U B , Hence q A > q B or, ( C A \u22c5 \u0394 T ) > ( C B \u22c5 \u0394 T ) \u21d2 C A > C B
Answer: A
\nOriginal PDF solution page
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Solution: q = n \u22c5 C m \u22c5 \u0394 T = 1 \u00d7 (0.22 \u00d7 32) \u00d7 (273 \u00d7 1.1 \u2013 273) \u00d7 4.2 J
Answer: C
\nOriginal PDF solution page
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Solution: q q n C T n C T V P V m P m = \u22c5 \u22c5 \u0394 \u22c5 \u22c5 \u0394 = , , 1 \u03b3 \u21d2 q V = \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 1 74 3 66 793 660 . J ( C P m , . = \u00d7 = 743 5 2 74 3 \u21d2 C V m , = 74.3 \u2013 8.3 = 66.0)
Answer: D
\nOriginal PDF solution page
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Solution: Isothermal: P.V = P V n P n P i i \u00d7 \u21d2 = \u22c5 Adiabatic: P.V r = P V n P n P a a \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u22c5 \u03b3 \u03b3 \u2234 P P n P n P n i a = \u22c5 \u22c5 = \u2212 \u03b3 \u03b3 1
Answer: C
\nOriginal PDF solution page
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Solution: \u0394 U = n C T T n R P V nR P V nR PV V m \u22c5 \u22c5 \u2212 = \u22c5 \u2212 \u22c5 \u22c5 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 , ( ) 2 1 1 2 1 \u03b3 \u03b3
Answer: A
\nOriginal PDF solution page
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Solution: K.E. = \u0394 U \u21d2 1 2 40 1000 100 8 314 1 5 1 2 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u00d7 \u0394 ( ) ( ) . ( . ) n n T \u2234 \u0394 T = 12.03 K
Answer: B
\nOriginal PDF solution page
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Solution: For minimum pressure, compression should be irreversible. \u0394 = \u21d2 \u22c5 \u2212 \u22c5 \u2212 = \u2212 \u2212 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f U w n R T T P V V P nRT P nRT P ext \u03b3 1 2 1 2 1 2 2 2 1 1 ( ) ( )\n4.38 Chapter 4 HINTS AND EXPLANATIONS or, T T T T P P P 2 1 2 2 2 1 2 1 700 400 1 4 1 700 400 100 \u2212 \u2212 = \u2212 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 \u2212 \u2212 = \u2212 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u03b3 . \u23a0 \u23a0 \u239f \u2234 P 2 = 362.5 kPa
Answer: C
\nOriginal PDF solution page
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Solution: q n C T n C T V m Ne V m SO = \u22c5 \u22c5 \u0394 + \u22c5 \u22c5 \u0394 ( ) ( ) , , 3 or, 12 \u00d7 10 3 = 2 \u00d7 3 \u00d7 ( T f \u2013 300) + 3 \u00d7 6 \u00d7 ( T f \u2013 400) \u21d2 T f = 875 K Now, P nRT V final atm = = \u00d7 \u00d7 = 5 0 08 875 10 35 .
Answer: A
\nOriginal PDF solution page
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Solution: Free expansion is isothermal.
Answer: B
\nOriginal PDF solution page
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Solution: C R R N R N R V m , ( ) ( ) = \u00d7 + \u00d7 + \u2212 \u00d7 = \u2212 3 1 2 3 1 2 3 6 3 3 \u2234 \u03b3 = = + = + \u2212 C C R C N P V V m 1 1 1 3 3 ,
Answer: C
\nOriginal PDF solution page
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Solution: C C P dV dT C P C RT V C R T V V m V m V m V m V m = + \u22c5 = + = + \u22c5 = + + , , , , ( ) \u03b1 \u03b1 \u03b1 \u03b1 0 = + C RT V P m , 0 \u03b1 Now, q C dT C dV C RT V dV m T T m V V P m V V = \u22c5 = \u22c5 \u22c5 = \u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u222b \u222b \u222b 1 2 1 2 1 2 0 ( ) , \u03b1 \u03b1 = \u22c5 \u2212 + \u22c5 \u03b1 C V V RT V V P m , ( ) ln 2 1 0 2 1
Answer: D
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Solution: w = \u201325 J = \u2013 nR \u22c5 \u0394 T \u0394 = \u22c5 \u22c5 \u0394 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u0394 = U n C T n R T J V m , 6 2 75 \u2234 q = \u0394 U \u2013 w = 100 J
Answer: B
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Solution: C C R x R R R m V m = + \u2212 = + \u2212 = , 1 3 2 1 5 2 5 6 \u2234 q n C T R J m = \u22c5 \u22c5 \u0394 = \u00d7 \u00d7 = \u22c5 1 5 6 26 180 14
Answer: C
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Solution: PV K dP dV x P V x x x = \u21d2 = \u2212 \u22c5 \u21d2 \u2212 = \u2212 \u00d7 \u21d2 = 1 4 2 1 2 \u2234 C C R x R R R m V m = + \u2212 = + \u2212 = , . 1 3 2 1 1 2 3 5
Answer: C
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Solution: V 0 2 V 0 V P 2 1 3
Answer: B
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Solution: V 1 V 2 V Isobaric Isothermal Adiabatic P \u0394 E adiabatic = Negative \u0394 E isothermal = 0 \u0394 E isobaric = Positive
Answer: B
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Solution: Boyle temperature, T B = 20 + 273 = 293 K Inversion temperature, T i = 2 \u00d7 T B = 586 K = 313\u00b0 C > 50\u00b0 C
Answer: C
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Solution: V 0 Mono Initial Di V 0 V 1 Mono Di V 0 3 4 Monoatomic : P V P V 1 0 5 3 2 1 5 2 \u22c5 = \u22c5 / / Diatomic : P V P V 1 0 7 5 2 0 7 5 3 4 \u22c5 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f / / \u2234 V V 1 0 21 25 3 4 = \u239b \u239d \u239c \u239e \u23a0 \u239f\n4.39 Thermodynamics HINTS AND EXPLANATIONS
Answer: B
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Solution: V 0 2 V V P P 2 V = Constant Isothermal reversible Adiabatic irreversible Adiabatic reversible
Answer: A
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Solution: For greater heat exchange, heat capacity should be high. C C R x m V m = + \u2212 , 1 for PV x = Constant
Answer: A
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Solution: q ABC = 600 + 200 = 800 J w AB = 0 and w N m m BC = \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 \u00d7 \u2212 \u2212 8 10 5 10 2 10 4 2 3 3 2 ( ) = \u2013240 J \u2234 \u0394 U AC = \u0394 U ABC = q ABC + w ABC = 800 + (\u2013240) = 560 J
Answer: C
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Solution: 3 60\u00b0 30\u00b0 A B C V 0 V 0 6 V 0 V 9 4 P 0 P 0 AB P V C : = + 3 1 P V C 0 0 1 3 = + (1) and 3 3 0 1 P V C B = + (2) BC P V C : = \u2212 + 1 3 2 P V C 0 0 2 1 3 6 = \u2212 \u22c5 + (3) 3 1 3 0 2 P V C B = \u2212 \u22c5 + (4) From equation (1), (2), (3) and (4), V V B = 9 4 0 Now, T T P V P V B A = \u22c5 \u22c5 = 3 9 4 27 4 0 0 0 0
Answer: A
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Solution: F = P 0 = P P 0 P i P f dw = F \u22c5 dx = ( P 0 \u2013 P ) A \u22c5 dx = ( P 0 \u2013 P ) \u22c5 dV \u2234 w P nRT V dV P V V RT V V V V = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u22c5 \u22c5 \u22c5 \u222b 0 0 \u03b7 \u03b7 \u03b7 ( ) ln P 0 V ( \u03b7 -1) \u2212 RT.ln \u03b7 = RT [ \u03b7 -1-ln \u03b7 ]
Answer: B
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Solution: T 0 ( V 0) P 0, P 0, P 1, P 2, T 0 \u22c5 ( V ) \u03b7 ( V 0) ( V ) Work performed on the piston = \u2212 \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u22c5 + \u222b \u222b \u22c5 P P dV dV V V V V 1 2 0 0 \u03b7 and ( V + h \u22c5 V ) = 2 V 0 = \u22c5 + P V 0 0 2 1 4 ln ( ) \u03b7 \u03b7
Answer: C
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Solution: Isothermal : P A \u22c5 V = P \u22c5 (2 V ) \u21d2 P A = 2 P Adiabatic : P B \u22c5 V 1.5 = P \u22c5 (2 V ) 1.5 \u21d2 P A = 2 2 P Isobaric : P C = P \u2234 P A : P B : P C = 2 : 2 2 : 1
Answer: B
\nOriginal PDF solution page
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Solution: A = ( V 0, T 0) , T 0 ( V 0, T 0) 27 8 = 27 8 P 0 , P 0 P 0 P 0 P f P f Chamber B P V P V B : 0 0 0 27 8 \u22c5 = \u22c5 \u03b3 \u03b3\n4.40 Chapter 4 HINTS AND EXPLANATIONS \u2234 V V T T B B = \u21d2 = 4 9 3 2 0 0 and V V V V T T A A = \u2212 = \u21d2 = 2 4 9 14 9 21 4 0 0 0 0 Now, q U U n C T T n C T T A A B V m A V m B = \u0394 + \u0394 = \u22c5 \u22c5 \u2212 + \u22c5 \u22c5 \u2212 , , ( ) ( ) 0 0 = \u22c5 \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u22c5 P V R T R T T T T P V 0 0 0 0 0 0 0 0 0 2 21 4 3 2 19 2
Answer: A
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Solution: h 1st step Final position x = ? mg A mg A P 0 P 1 After 1st step, the process is irreversible adiabatic. Hence, \u0394 U = w n C T T P V V V m \u22c5 \u22c5 \u2212 = \u2212 \u2212 , ( ) ( ) 2 1 2 1 ext or, n R P V nR PV nR P V V A H \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 + \u22c5 3 2 1 2 1 1 1 2 1 [ ( )] \u2234 V 2 = V 1 + 0.4 H.A \u21d2 x = 0.4 H (The final pressure of gas after 2nd step will remain same as initial, beginning of processes.)
Answer: A
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Solution: Smaller the heat capacity larger is \u0394 T.
Answer: B
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Solution: \u0394 = \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = H n C T T P m , ( ) ( ) 2 1 1 40 500 300 8000 J \u2234 \u0394 U = \u0394 H \u2013 P \u22c5 \u0394 V = 8000 \u2013 2(40 \u2013 30) \u00d7 100 = 6000 J
Answer: C
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Solution: q = 0 \u21d2 \u0394 U = w = \u2013 P ext \u22c5 ( V 2 \u2013 V 1 ) = \u20134 \u00d7 (30 \u2013 40) = 40 l -bar Now, \u0394 H = \u0394 U + \u0394 ( PV ) = 40 + (4 \u00d7 30 \u2013 2 \u00d7 40) = 80 L -atom = 8000 J
Answer: C
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Solution: \u0394 U = 0 \u0394 H = \u0394 U + \u0394 ( PV ) = 0 + B ( P 2 \u2013 P 1 ) = B RT V B RT V B \u22c5 \u2212 \u2212 \u2212 2 1 = \u00d7 \u00d7 \u2212 \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 2 8 314 400 1 22 2 1 12 2 332 56 . . J
Answer: C
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Solution: \u0394 = \u22c5 \u22c5 \u2212 U n C T T V m 1 2 1 , ( ) \u0394 H 1 = \u0394 U 1 + V \u22c5 \u0394 P = 1 \u00d7 C V,m \u00d7 ( T 2 \u2013 T 1 ) + V 1 ( P 2 \u2013 P 1 ) Now, \u0394 U 2 = w 2 = \u2013 P ext ( V 2 \u2013 V 1 ) = \u2013 P 3 ( V 2 \u2013 V 1 ) \u0394 H 2 = \u0394 U 2 + \u0394 ( PV ) = \u2013 P 3 ( V 2 \u2013 V 1 ) + ( P 3 V 2 \u2013 P 2 V 1 ) \u2234 \u0394 H total = \u0394 H 1 + \u0394 H 2 = C V ( T 2 \u2013 T 1 ) + V 1 ( P 3 \u2013 P 1 )
Answer: A
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Solution: \u03b7 = \u2212 1 T T C H 1 6 1 = \u2212 T T C H and 1 3 1 65 390 = \u2212 \u2212 \u21d2 = T T T C H H K = 117\u00b0 C
Answer: C
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Solution: q q T T q C H rej abs rej cal = \u21d2 = \u00d7 = 390 600 120 78
Answer: B
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Solution: 1 1 \u2212 \u2212 \u0394 \u239b \u239d \u239c \u239e \u23a0 \u239f > \u2212 + \u0394 \u239b \u239d \u239c \u239e \u23a0 \u239f T T T T T T C H C H
Answer: A
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Solution: T V T V H C \u22c5 = \u22c5 \u2212 \u2212 2 1 3 1 \u03b3 \u03b3 \u21d2 T T V V C M = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 2 3 1 1 4 1 1 2 75 1 1 5 \u03b3 . . . \u2234 \u03b7 = \u2212 = 1 1 3 T T C H
Answer: C
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Solution: 2 1 3 4 V T 2 T 3 T 4 T 1 P T V T V 2 1 1 3 2 1 \u22c5 = \u22c5 \u2212 \u2212 \u03b3 \u03b3 and T V T V 1 1 1 4 2 1 \u22c5 = \u22c5 \u2212 \u2212 \u03b3 \u03b3 \u2234 T T T T T T T T T T 2 1 3 4 2 1 1 3 4 4 = \u21d2 \u2212 = \u2212 @TheBookCorner\n4.41 Thermodynamics HINTS AND EXPLANATIONS \u03b7 = \u2212 = \u2212 \u22c5 \u2212 \u22c5 \u2212 = \u2212 = \u2212 \u239b \u239d \u239c 1 1 1 1 3 4 2 1 4 1 1 2 q q n C T T n C T T T T V V V m V m rej abs , , ( ) ( ) \u239e \u239e \u23a0 \u239f \u2212 \u03b3 1 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 1 1 10 0 6 7 5 1 .
Answer: C
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Solution: \u0394 S unit = \u0394 S Source + \u0394 S Heat engine + \u0394 S Sink = \u2212 \u00d7 + + \u00d7 = + 40 10 500 0 30 10 300 20 3 3 J/K
Answer: B
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Solution: \u0394 = \u22c5 \u22c5 S n C T T P m , ln 2 1 = \u00d7 \u00d7 = \u2212 3 2 32 14 900 1000 0 14 . ln . c al/K
Answer: A
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Solution: \u0394 = \u22c5 S nR V V ln 2 1 = \u00d7 \u00d7 = 2 8 314 2 34 58 3 3 . ln ( ) . a a J/K
Answer: B
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Solution: \u0394 = \u22c5 \u22c5 S n C T T P m , ln 2 1 = \u00d7 \u00d7 = 1 5 2 1000 250 7 0 R ln . cal/K
Answer: C
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Solution: \u0394 = \u22c5 \u22c5 S n C T T V m , ln 2 1 S R K 500 46 2 1 3 2 500 250 \u2212 = \u00d7 \u00d7 . ln \u2234 S 500 K = 48.3 Ccal/K-mol
Answer: A
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Solution: \u0394 = \u22c5 S nR V V ln 2 1 \u21d2 \u2212 = \u00d7 \u00d7 \u00d7 \u21d2 = \u2212 5 0 10 15 10 300 15 5 4 5 3 2 2 . ( ) ln . V V L
Answer: B
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Solution: \u0394 = \u2212 \u00d7 = \u2212 S Surr J/K 1 5 10 300 5 3 . Now, \u0394 S unit = \u0394 S Sys + \u0394 S Surr = 5.51 + (\u20135) = + 0.51 J/K Hence, the process is irreversible.
Answer: C
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Solution: \u0394 = \u22c5 \u22c5 + \u22c5 S n C T T nR P P P m , ln ln 2 1 1 2 or, 0 5 2 1200 300 1 32 2 2 = \u00d7 \u00d7 + \u00d7 \u21d2 = n R nR P P ln ln bar
Answer: C
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Solution: \u0394 S = \u0394 S adiabatic + \u0394 S isobaric = 0 2 1 + \u22c5 \u22c5 n C T T P m , ln = \u00d7 \u00d7 = \u2212 1 6 4 5 2 1 3 2 2 . ln . R cal/K
Answer: A
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Solution: \u0394 = \u22c5 \u22c5 + \u22c5 S n C T T nR V V V m , ln ln 2 1 2 1 = \u22c5 \u22c5 + \u22c5 \u22c5 n C P P n C V V V m P m , , ln ln 2 1 2 1 = \u00d7 \u2212 \u00d7 + \u00d7 \u2212 \u00d7 2 1 5 1 1 4 2 1 5 1 5 1 2 R R . ln . . ln = \u201311.64 J/K
Answer: C
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Solution: S S n C T T nR V V V m 2 1 2 1 2 1 \u2212 = \u22c5 \u22c5 + \u22c5 , ln ln = \u00d7 \u00d7 + \u00d7 \u00d7 1 2 3 2 1 2 1 2 2 . l n . ln R R = \u20130.84 cal/K
Answer: A
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Solution: \u0394 = \u22c5 \u22c5 + \u22c5 S n C T T nR V V V m , ln ln 2 1 2 1 = \u00d7 \u2212 \u00d7 + \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 1 1 1 2 1 1 2 1 1 R T T R T T n \u03b3 ln ln / (as T.V n -1 = Constant) R T T n \u22c5 \u2212 \u2212 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ln 2 1 1 1 1 1 \u03b3 = \u2212 \u2212 \u2212 \u22c5 ( ) ( )( ) ln n R n \u03b3 \u03b3 \u03c4 1 1
Answer: A
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Solution: \u0394 = \u22c5 \u22c5 + \u22c5 \u22c5 S n C P P n C V V V m P m , , ln ln 2 1 2 1 = \u00d7 \u00d7 + \u00d7 \u00d7 2 3 2 2 2 5 2 2 R R ln ln = + 11.2 cal/K
Answer: B
\nOriginal PDF solution page
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Solution: dS n C dT P dV T V m = \u22c5 \u22c5 + \u22c5 , For maximum entropy, dS dV = 0\n4.42 Chapter 4 HINTS AND EXPLANATIONS or, n C dT dV P V m \u22c5 + = , 0 (1) Now, P RT V P V dT dV R P V = = \u2212 \u21d2 = \u2212 0 0 1 2 \u03b1 \u03b1 ( ) (2) From (1) and (2), 1 1 1 2 0 0 0 \u00d7 \u2212 \u00d7 \u2212 + \u2212 = R R P V P V \u03b3 \u03b1 \u03b1 ( ) ( ) \u2234 V P = \u22c5 + \u03b3 \u03b1 \u03b3 0 1 ( )
Answer: C
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Solution: dS C dT T P dV T a dT C T dT V m V m = \u22c5 + \u22c5 = \u22c5 + \u22c5 \u22c5 , , 1 or, R V dV a dT R V V a T T V V T T 0 0 0 0 \u222b \u222b \u22c5 = \u22c5 \u21d2 \u22c5 = \u2212 ln ( ) \u2234 T = T 0 + R a V V \u22c5 ln 0
Answer: A
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Solution: dS C dT T T dT T S aT S T T 0 0 3 3 0 3 \u222b \u222b \u222b = \u22c5 = \u22c5 \u21d2 =
Answer: B
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Solution: \u0394 = \u0394 + \u0394 = \u2212 + = + S S S A B 12000 600 12000 400 10 J/K
Answer: C
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Solution: Heat lost by alloy = Heat gained by water or 4 \u00d7 4 \u00d7 (800 \u2013 T ) = 4 \u00d7 1.0 \u00d7 ( T \u2013 300) \u21d2 T = 700 K (As date is not given for vaporization of water) Now, \u0394 S mix = \u0394 S alloy + \u0394 S water = 4 \u00d7 4 \u00d7 ln 700 800 + 4 \u00d7 1 \u00d7 ln 700 300 = 1.0 K cal/K
Answer: A
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Solution: Final temperature of both blocks = T T 1 2 2 + \u2234 \u0394 S = \u0394 S 1 + \u0394 S 2 = C T T T C T T T \u22c5 + + \u22c5 + ln ( ) / ln ( ) / 1 2 1 1 2 2 2 2 = \u22c5 + C T T T T ln ( ) 1 2 2 1 2 4
Answer: A
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Solution: \u0394 S = \u2013 R [ n 1 \u22c5 ln x 1 + n 2 \u22c5 ln x 2 ] = \u2013 R [0.8 \u00d7 ln 0.8 + 0.2 \u00d7 ln 0.2] = + 0.96 Cal/K.
Answer: D
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Solution: Larger molar mass, greater is the molar entropy.
Answer: C
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Solution: Greater the number of atoms, greater is the molar entropy.
Answer: C
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Solution: nC( s ) + (n + 1) H 2 ( g ) \u2192 C n H 2n + 2 ( g ) with increase in n , the decrease in entropy increases.
Answer: C
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Solution: H 2 O ( l , 1 atm, 100\u00b0C) ( ) 1 \u23af \u2192 \u23af H 2 O ( g , 1 atm, 100\u00b0C) ( ) 2 \u23af \u2192 \u23af H 2 O ( g , 5 atm, 100\u00b0C) \u0394 G 1 = 0 and \u0394 G 2 = nRT ln P P 2 1 = 5 \u00d7 2 \u00d7 373 \u00d7 ln 5 1 = 3730 ln 5 Cal
Answer: B
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Solution: q = \u0394 U \u2013 w = 0 \u2013 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f nRT P P ln 1 2 = nRT P P \u22c5 ln 1 2 = \u2013 \u0394 G \u2234 \u0394 G = \u2013 q = \u2013(\u20131200) = +1200 cal
Answer: C
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Solution: \u0394 G \u00b0 = \u2013 R T \u22c5 ln K eq \u21d2 \u20131743 = \u2013 8.3 \u00d7 300 \u00d7 ln K eq \u2234 K eq = 2 Now, (a) K eq = \u00d7 3 3 6 (b) K eq = \u00d7 6 3 3 2 (c) K eq = \u00d7 6 3 3 2 (d) K eq = \u00d7 3 3 6 2
Answer: B
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Solution: H O C Pa H O C Pa H O 2 2 2 ( , , . ) ( , , . ) ( l s G G g \u2212 \u00b0 \u23af \u2192 \u23af \u2212 \u00b0 \u2193 \u0394 = \u2191 \u0394 = 10 0 28 10 0 26 0 0 1 3 , , , . ) ( , , . ) \u2212 \u00b0 \u23af \u2192 \u23af\u23af \u2212 \u00b0 \u0394 10 0 28 10 0 26 2 C Pa H O C Pa 2 G g \u0394 G 2 = nRT ln P P 2 1 = 1 \u00d7 R \u00d7 263 \u00d7 ln 0 26 0 28 . . \u2234 \u0394 G = \u0394 G 1 + \u0394 G 2 + \u0394 G 3 = 263 R ln 13 14
Answer: D
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Solution: Free expansion is isothermal \u0394 G = n RT ln P P 2 1 = nRT ln V V 2 1 = 10 5 \u00d7 (1.2 \u00d7 10 \u20133 ) \u00d7 ln 1 2 2 4 . . = \u201384 J\n4.43 Thermodynamics HINTS AND EXPLANATIONS
Answer: D
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Solution: \u0394 G 1 \u00b0 = \u2013 RT \u22c5 ln K 1 and \u0394 G 2 \u00b0 = \u2013 RT \u22c5 ln K 2 Now, \u0394 G 2 \u00b0 \u2013 \u0394 G 1 \u00b0 = \u2013 RT [ln K 2 \u2013 ln K 1 ] = \u2013 RT [ln e 4 ] = \u20132 \u00d7 300 \u00d7 4 = \u20132400 cal
Answer: A
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Solution: At 0.04 atom, the system is in equilibrium.
Answer: B, C, D
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Answer: A, B, C
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Answer: A, B, C
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Solution: For isolated system, there should not be any mass and energy transfer with surroundings.
Answer: A, B, D
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Solution: 2 V V V Given 0.5 P Isothermal P P 0.5 P T P P
Answer: A, B
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Solution: The internal energy of real gas may change on changing the volume of gas. Change in physical state also changes the physical state.
Answer: A, B, C
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Answer: C
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Answer: A, B, C, D
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Solution: Option (c) should be changed with (c) adiabatic free expansion of any gas is also isothermal.
Answer: A, C
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Solution: w rev \u2013 w irr = (\u2013 P \u22c5 dV ) \u2013 (\u2013 P ext \u22c5 dV ) = ( P ext \u2013 P ) \u22c5 dV = negative, always and q rev \u2013 q irr = positive, always
Answer: A, C
\nOriginal PDF solution page
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Solution: q = 0 w = \u0394 U = n C T T R V m \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = \u2212 , ( ) ( ) cal 2 1 4 3 2 290 320 360 \u0394 H = \u03b3 \u22c5 \u0394 U = 5 3 360 600 \u00d7 \u2212 = \u2212 ( ) cal
Answer: A, B, C
\nOriginal PDF solution page
Open page 228 in PDFExtracted text
Answer: A, B, C
\nOriginal PDF solution page
Open page 228 in PDFExtracted text
Solution: Process BC : P T P T P P B B C C B B = \u21d2 = \u21d2 = 500 1 250 2 bar and \u0394 = \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = \u2212 U n C T T BC V m C B 1 2 1 5 250 500 750 ( ) . R ( ) R Process CD : \u0394 = \u21d2 \u22c5 \u2212 = \u2212 \u2212 U w n C T T P V V V m D C D C 1 ( ) ( ) ext or, n R T T P nRT P nRT P T D C D D D C C D \u00d7 \u00d7 \u2212 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = 1 5 450 . ( ) K and \u0394 = \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = H n C T T R CD P m D C 1 2 2 5 450 250 1000 ( ) . ( ) R
Answer: C, D
\nOriginal PDF solution page
Open page 228 in PDFExtracted text
Solution: PV K P K V dP dV K V P V \u03b3 \u03b3 \u03b3 \u03b3 \u03b3 = \u21d2 = \u22c5 \u21d2 = \u22c5 \u2212 = \u2212 \u22c5 \u2212 \u2212 \u2212 1 1 1 1 ( ). The gas having higher \u03b3 will have higher magnitude of slope of P vs. V curve. Now, n R dT P dV nRT V dV \u22c5 \u2212 \u22c5 = \u2212 \u22c5 = \u2212 \u22c5 \u03b3 1 or, dV dT V T = \u2212 \u2212 \u22c5 1 1 \u03b3\n4.44 Chapter 4 HINTS AND EXPLANATIONS Gas having higher \u03b3 will have lower magnitude of slope of V vs. T curve. Now, n C dT P dV nRdT V dP V m \u22c5 \u22c5 = \u2212 \u22c5 = \u2212 \u2212 \u22c5 , [ ] or, n C dT V dP n R dT nRT P dP P m \u22c5 \u22c5 = \u2212 \u22c5 \u21d2 \u22c5 \u2212 \u22c5 = \u22c5 , \u03b3 \u03b3 1 \u2234 dP dT P T = \u2212 \u22c5 \u03b3 \u03b3 1 Gas having higher \u03b3 will have lower magnitude of slope of P vs. T curve.
Answer: C
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Answer: A
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Answer: A, B, C
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: \u03b3 V V 2 V CH4 SO2 O2 Ne P Here, \u03b3 decreases on increasing degree of freedoms. As fi nal pressure is minimum for Ne, its fi nal temperature is minimum (decrease in temperature is maximum). Now, for overall process, \u0394 T = 0 \u21d2 \u0394 U total = 0 or, \u0394 U I + \u0394 U II = 0 \u21d2 (0 + w I ) + ( q II + 0) = 0 \u2234 q II = \u2013 w I = maximum for CH 4
Answer: A, B
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Answer: B, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: C C R P m V m , , \u2212 = \u21d2 S S R M P V \u2212 = = 0 04545 . \u2234 M = 44 gm/mol
Answer: C, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: q = 0 \u21d2 \u0394 U = w = \u2013 P 0 (4 V 0 \u2013 V 0 ) = \u20133 P 0 V 0 Now, \u0394 H = \u0394 U + \u0394 ( PV ) = (\u20133 P 0 V 0 ) + ( P 0 \u22c5 4 V 0 \u2013 2 P 0 \u22c5 V 0 ) = \u2013 P 0 \u22c5 V 0
Answer: B, C, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Answer: A, B, C, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Answer: A, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: In reversible cycle, heat rejected is minimum. For reversible cycle, q T T q C H rej abs J = \u00d7 = \u00d7 = 400 500 100 80
Answer: A, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: dS q T = rev and \u001e \u222b dS = 0
Answer: A, B, C
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: In rusting, moles of gas decreases.
Answer: B, C
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Answer: B, C, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: \u0394 U = 0 \u21d2 q = \u2013 w
Answer: B, C, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: For a process to be spontaneous at low temperature and non-spontaneous at high temperature, \u0394 H = negative and \u0394 S = negative.
Answer: A, B, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: ( ) , \u0394 = \u00d7 = S J K K Vap atm mol 350 1 3 35 10 350 100 On increasing pressure at constant temperature entropy decreases. ( ) , \u0394 = G K Vap atm 350 1 0 On increasing pressure at constant temperature energy increases.
Answer: B, D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Comprehension 1 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: \u0394 = \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u2212 \u2212 = \u2212 U n C T T R V m 1 2 1 4 5 2 50 0 1000 ( ) ( ) cal
Comprehension 1 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: \u0394 H = \u03b3 \u22c5 \u0394 U = 7 5 1000 1400 \u00d7 \u2212 = \u2212 ( ) cal
Comprehension 1 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 229 in PDFExtracted text
Solution: w = 0 ( V = Constant)\n4.45 Thermodynamics HINTS AND EXPLANATIONS Comprehension II
Comprehension 2 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: q U n C T n C T C C V m V m m V m = \u2212\u0394 \u21d2 \u22c5 \u22c5 \u0394 = \u2212 \u22c5 \u22c5 \u0394 \u21d2 = \u2212 , , ,
Comprehension 2 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: C C R x m V m = + \u2212 , 1 \u21d2 \u2212 = + \u2212 \u21d2 \u2212 \u22c5 \u2212 = \u2212 C C R x R R x V m V m , , 1 2 1 1 \u03b3 \u2234 x = + \u03b3 1 2 Now, T \u22c5 V x \u20131 = Constant \u21d2 T \u22c5 V ( \u03b3 \u20131)/2 = Constant
Comprehension 2 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: T T V V T T 2 1 1 2 1 2 2 5 3 1 2 2 300 1 8 150 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u03b3 / / K \u2234 w nR T T x = \u2212 \u2212 \u2212 = \u2212 \u00d7 \u2212 \u2212 + \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 ( ) ( ) / 2 1 1 1 2 150 300 1 5 3 1 2 900 cal Comprehension III
Comprehension 3 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: w = \u2013 nR \u22c5 \u0394 T = \u20131 \u00d7 8.314 \u00d7 72 = \u2013598.6 J = \u2013 0.6 kJ
Comprehension 3 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: \u0394 U = q + w = 1.6 + (\u20130.6) = 1.0 kJ
Comprehension 3 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: \u03b3 = \u0394 \u0394 = = H U 1 6 1 0 1 6 . . . Comprehension IV
Comprehension 4 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "D"
+ ],
+ "answer": null,
+ "explanation": "Answer: D
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: V 2 = 4 V 0 \u21d2 P 2 = 4 P 0 As PV T P V T 1 1 1 2 2 2 = \u21d2 P V T P P T 0 0 0 0 0 2 4 4 = \u22c5 \u21d2 T T 2 0 16 = Now, \u0394 = \u22c5 \u22c5 \u0394 = \u00d7 \u2212 \u00d7 \u2212 = \u2212 = \u2212 U n C T n R T T P V V V m , ( . ) \u03b3 \u03b3 \u03b1 \u03b3 1 16 15 1 15 1 0 0 0 0 0 2
Comprehension 4 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: w nR T T x nR T V = \u2212 \u2212 = \u00d7 \u2212 \u2212 = ( ) ( ) 2 1 0 0 2 1 15 1 1 15 2 \u03b1
Comprehension 4 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: C C R x R R R m V m = + \u2212 = \u2212 + \u2212 \u2212 = + \u2212 1 1 1 1 1 1 2 1 \u03b3 \u03b3 \u03b3 ( ) ( ) ( ) Comprehension V
Comprehension 5 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: P = a \u22c5 T a = a \u22c5 PV nR \u239b \u239d \u239c \u239e \u23a0 \u239f \u03b1 \u21d2 P V \u22c5 = \u2212 \u03b1 \u03b1 1 Constant \u2234 w nR T x R T R T = \u22c5 \u0394 \u2212 = \u00d7 \u22c5 \u0394 \u2212 \u2212 = \u2212 \u22c5 \u0394 1 1 1 1 1 \u03b1 \u03b1 \u03b1 ( )
Comprehension 5 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: C C R x R R R m V m = + \u2212 = \u2212 + \u2212 \u2212 = \u2212 + \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 , ( ) . 1 1 1 1 1 1 1 \u03b3 \u03b1 \u03b1 \u03b3 \u03b1
Comprehension 5 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 230 in PDFExtracted text
Solution: 1 1 1 0 \u03b3 \u03b1 \u2212 + \u2212 < ( ) \u21d2 \u03b1 \u03b3 \u03b3 > \u2212 1\n4.46 Chapter 4 HINTS AND EXPLANATIONS Comprehension VI
Comprehension 6 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: U V n C T V m = \u22c5 = \u22c5 \u22c5 \u03b1 \u03b1 , \u21d2 T V \u22c5 = \u2212 \u03b1 Constant As T V x \u22c5 = \u2212 1 Constant \u21d2 x \u2013 1 = \u2013 a Now, w n R T x U = \u22c5 \u22c5 \u0394 \u2212 = \u2212 \u22c5 \u0394 \u2212 1 1 ( ) \u03b3 \u03b1
Comprehension 6 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: q U w U U U = \u0394 \u2212 = \u0394 + \u2212 \u22c5 \u0394 = \u0394 + \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ( ) \u03b3 \u03b1 \u03b3 \u03b1 1 1 1
Comprehension 6 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: C R R x R R m = \u2212 + \u2212 = \u2212 + \u03b3 \u03b3 \u03b1 1 1 1 Comprehension VII
Comprehension 7 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "B"
+ ],
+ "answer": null,
+ "explanation": "Answer: B
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: PV P V 1 1 2 2 \u03b3 \u03b3 = \u21d2 P 2 7 5 1 320 10 128 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = / atm
Comprehension 7 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: PV T P V T 1 1 1 2 2 2 = \u21d2 1 320 300 128 10 2 \u00d7 = \u00d7 T \u21d2 T 2 1200 = K
Comprehension 7 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: w n C T V m = \u22c5 \u22c5 \u0394 = \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 = , . . . ( ) . 1 0 32 0 082 300 5 2 8 314 1200 300 243 3 3 J Comprehension VIII
Comprehension 8 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: PV P V P 1 1 2 2 2 5 3 1 8 32 \u03b3 \u03b3 = \u21d2 = \u00d7 = ( ) / atm
Comprehension 8 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: For A : P 1 = 1 atm, P 2 = 32 atm V 1 = VL , V 2 = V + 7 8 V = 15 8 V L T 1 = 27.3 K; T 2 = ? Now, PV T P V T 1 1 1 2 2 2 = \u21d2 T 2 = 1638 K
Comprehension 8 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
Open page 231 in PDFExtracted text
Solution: \u0394 = \u22c5 \u22c5 \u0394 = \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 H n C T A P m , . . . ( . ) 1 22 4 0 082 27 3 5 2 2 1638 27 3 = 80535 cal Comprehension IX
Comprehension 9 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: \u0394 U = 0 \u0394 H = \u0394 U + V \u22c5 \u0394 P = 0.9 L \u00d7 1 532 760 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f atm = 0.27 L -atm = 27 J
Comprehension 9 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: \u0394 = \u2192 U 1 2 0 \u0394 = \u0394 \u2192 U U 2 3 for temperature increase + \u0394 U for vaporization of water. = m . s . \u22c5 \u0394 T + ( q + w ) = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 900 4 2 1000 20 450 18 40 450 18 8 1000 373 . \u239f \u239f = 1001 kJ
Comprehension 9 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: \u0394 = \u0394 + = \u00d7 + \u00d7 \u00d7 \u239b \u239d \u239c \u239c \u239c \u239e \u23a0 \u239f \u239f \u239f \u2192 \u2192 \u2192 H H q 1 3 1 2 2 3 27 450 18 40 4 2 1000 20 J+ kJ 900 kJ . = 1075.573 kJ
Comprehension 10 - subquestion 1
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: w 1 2 0 \u2192 = w 2 3 450 18 8 1000 373 74 6 \u2192 = \u2212 \u00d7 \u00d7 = \u2212 . kJ\n4.47 Thermodynamics HINTS AND EXPLANATIONS Comprehension X
Comprehension 10 - subquestion 2
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "A"
+ ],
+ "answer": null,
+ "explanation": "Answer: A
\nOriginal PDF solution page
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Solution: T A > T B < T C < T D = T A From question : T T A B = 4 and T A = 800 K \u21d2 T B = 200K Also, V A > V B > V C = V D From equation : V V A C = 8 2 and V V T T A B A B = = 4 For process BC : T V T V B B C C \u22c5 = \u22c5 \u2212 \u2212 \u03b3 \u03b3 1 1 \u21d2 T T V V C B B C r = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 1 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 200 8 2 4 400 5 3 1 K
Comprehension 10 - subquestion 3
",
+ "options": [
+ {
+ "identifier": "A",
+ "content": ""
+ },
+ {
+ "identifier": "B",
+ "content": ""
+ },
+ {
+ "identifier": "C",
+ "content": ""
+ },
+ {
+ "identifier": "D",
+ "content": ""
+ }
+ ],
+ "correct_options": [
+ "C"
+ ],
+ "answer": null,
+ "explanation": "Answer: C
\nOriginal PDF solution page
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Solution: \u0394 U BC = C T T R V m C B , ( ) ( ) . \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = 1 3 2 400 200 2 4942 kJ
Answer: B
\nOriginal PDF solution page
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Solution: q = 0 because \u0394 U = 0 and w = 0
Answer: C
\nOriginal PDF solution page
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Solution: Enthalpy of ideal gas is independent from pressure.
Answer: D
\nOriginal PDF solution page
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Solution: For non-ideal gas, U = f ( T , V )
Answer: D
\nOriginal PDF solution page
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Solution: In adiabatic free expansion, \u0394 T = 0
Answer: A
\nOriginal PDF solution page
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Solution: V 1 V 2 V Isothermal Adiabatic P P 1 Pi Pa
Answer: A
\nOriginal PDF solution page
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Solution: V 1 V 2 V Isothermal Adiabatic P Pa Pi P 2
Answer: A
\nOriginal PDF solution page
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Solution: Magnitude of work in adiabatic process depends on change in temperature.
Answer: A
\nOriginal PDF solution page
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Solution: Magnitude of work in adiabatic process depends on change in temperature.
Answer: A
\nOriginal PDF solution page
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Solution: q n C T P P m = \u22c5 \u22c5 \u0394 ,
Answer: C
\nOriginal PDF solution page
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Solution: Endothermic reactions may also be spontaneous.
Answer: A
\nOriginal PDF solution page
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Solution: At high temperature, process may become entropy driven.
Answer: A
\nOriginal PDF solution page
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Solution: At low temperature, process may become enthalpy driven.
Answer: C
\nOriginal PDF solution page
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Solution: If \u0394 G = negative and \u0394 S = negative, \u0394 H must be negative and \u0394 G = \u0394 H \u2013 T \u22c5 \u0394 S.
Answer: D
\nOriginal PDF solution page
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Answer: D
\nOriginal PDF solution page
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Solution: \u0394 S sys = 0 but \u0394 S univ = +ve\n4.48 Chapter 4 HINTS AND EXPLANATIONS
Answer: A \u2192 P, R, S, T; B \u2192 P, Q, R, S; C \u2192 Q, R, S, T; D \u2192 P, Q, R, T
\nOriginal PDF solution page
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Answer: A \u2192 Q; B \u2192 Q, S; C \u2192 Q, S; D \u2192 P
\nOriginal PDF solution page
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Answer: A \u2192 P, R, S; B \u2192 Q, R, S; C \u2192 Q, R, S; D \u2192 R, S
\nOriginal PDF solution page
Open page 233 in PDFExtracted text
Answer: A \u2192 P, Q, R, S; B \u2192 R, S; C \u2192 Q
\nOriginal PDF solution page
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Solution: For ideal gas, H = f ( T ) but in general, H = f ( T , P )
Answer: A \u2192 P, S; B \u2192 Q, R, S; C \u2192 R
\nOriginal PDF solution page
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Solution: U f T V dU U T dT U V dV V T = \u21d2 = \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 + \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 ( , ) \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 U T n C U V T P T P V V m T V , and
Answer: A \u2192 P, R; B \u2192 Q, S; C \u2192 R; D \u2192 Q, S
\nOriginal PDF solution page
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Solution: N 2 ( g ) + O 2 ( g ) \u2192 2NO ( g ) ; \u0394 H = Positive, \u0394 S \u2248 0 2 KI ( aq ) + HgI 2 ( aq ) \u2192 K 2 [HgI 4 ]( aq ) ; \u0394 H = Negative, \u0394 S \u2248 Negative PCl 5 ( g ) \u2192 PCl 3 ( g ) + Cl 2 ; \u0394 H = Positive, \u0394 S \u2248 Positive NH 3 ( g ) + HCl ( g ) \u2192 NH 9 Cl ( s ); \u0394 H = Negative, \u0394 S = Negative
Answer: A \u2192 Q; B \u2192 P, S; C \u2192 R, S
\nOriginal PDF solution page
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Solution: (A) \u0394 H = 0, \u0394 U = 0, \u0394 S total = 0, \u0394 S Sys = Positive (B) q = 0, \u0394 S Sys = 0, \u0394 S total = 0, \u0394 S surr = 0 (C) q = 0, \u0394 S surr = 0, \u0394 S total = Positive.
Answer: A \u2192 P, S; B \u2192 P, R, S; C \u2192 Q; D \u2192 R, S
\nOriginal PDF solution page
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Solution: (A) \u0394 H = Negative, \u0394 V = \u00b1 V e (B) \u0394 H = \u00b1 V e, \u0394 S = \u2212 V e, \u0394 G = \u0394 H \u2013 T , \u0394 S = + V e, if \u0394 H = + V e = \u00b1 V e, if \u0394 H = \u2212 V e (C) \u0394 H = Ea f \u2013 Ea b = 10 kJ / mol = + V e \u0394 S = + V e \u2234 \u0394 G = \u0394 H \u2013 T \u0394 S = \u2212 V e, at high temperature (D) \u0394 H = + V e, \u0394 S \u2248 0 \u21d2 \u0394 G \u2248 \u0394 H
Answer: A \u2192 P, S, R; B \u2192 P, R; C \u2192 P; D \u2192 Q, R, S
\nOriginal PDF solution page
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Solution: (A) Solid \u001f Liquid ; \u0394 G = 0, \u0394 S = Positive, \u0394 V \u2248 0 \u21d2 \u0394 H \u2248 \u0394 U (B) Liquid \u001f Vapour : \u0394 G = 0, \u0394 S = Positive, (C) Triple point is equilibrium condition. (D) Melting at boiling point is spontaneous.
Answer: A \u2192 S; B \u2192 P; C \u2192 Q; D \u2192 R
\nOriginal PDF solution page
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Solution: dG = V \u22c5 dP \u2013 S \u22c5 dT \u21d2 ( dG ) T = V \u22c5 dP and ( dG ) P = \u2013 S \u22c5 dT
Answer: 6
\nOriginal PDF solution page
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Solution: w = \u2013 P ext ( V 2 \u2013 V 1 ) = \u2013 P ext [ V 0 (1 + \u03b3 \u22c5 t 2 ) \u2013 V 0 (1 + \u03b3 \u22c5 t 1 )] = \u2013 P ext \u22c5 V 0 \u22c5 \u03b3 \u22c5 ( t 2 \u2013 t 1 ) \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 \u00b0 \u00d7 \u00b0 \u2212 10 18 10 0 0002 10 5 2 6 3 N m ( ) . m C C = 0.0036 J
Answer: 6
\nOriginal PDF solution page
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Solution: q = \u0394 U \u2013 w = 0 \u2013 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f nRT V V ln 2 1 or, 420 = 1 \u00d7 2 \u00d7 300 \u00d7 ln V 2 1 0 082 300 8 21 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f . . \u21d2 V 2 = 6 L
Answer: 1
\nOriginal PDF solution page
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Solution: P = K \u22c5 l For initial condition, 1 bar = K \u00d7 1 m \u21d2 K = 1 bar/m And, V r l = = 4 3 6 3 3 \u03c0 \u03c0 \u21d2 dV l dl = \u22c5 \u03c0 2 2 Now, w P dV K l l dl K l l l l V V = \u2212 \u22c5 = \u2212 \u22c5 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u22c5 \u2212 \u222b \u222b ( ) \u03c0 \u03c0 2 2 4 2 2 4 1 4 1 2 1 2 = \u2212 \u00d7 \u00d7 \u2212 \u2212 \u00d7 10 2 4 1 4 1 10 5 2 4 4 4 7 N/m m J \u03c0 ( ) m \u001c
Answer: 6
\nOriginal PDF solution page
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Solution: \u2013 w = mgh \u21d2 P ext ( V 2 \u2013 V 1 ) = mgh = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 \u00d7 = \u00d7 \u00d7 \u21d2 = \u2212 4 10 8 2 10 40 10 6 5 2 3 3 N m m ( ) m h h
Answer: 6
\nOriginal PDF solution page
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Solution: q U n C T n C T Q Q Q m V m \u0394 = \u22c5 \u22c5 \u0394 \u22c5 \u22c5 \u0394 = \u2212 , 2 \u21d2 C R m 3 2 2 = \u21d2 C m = 6 cal/K mole\n4.49 Thermodynamics HINTS AND EXPLANATIONS
Answer: 7
\nOriginal PDF solution page
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Solution: q w n C T nR T P m = \u22c5 \u22c5 \u0394 \u2212 \u22c5 \u0394 , \u21d2 q R R \u2212 = \u2212 2 7 2 \u21d2 q = 7 J
Answer: 3
\nOriginal PDF solution page
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Solution: q = q 1 + q 2 = \u0394 U 1 + \u0394 H 2 = n C n C V m P m \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f , , 300 2 300 300 300 2 = \u22c5 \u22c5 = \u00d7 \u00d7 = n R 300 2 10 2 300 2 3000 cal
Answer: 5
\nOriginal PDF solution page
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Solution: State I State II Isothermal Isochoric \u23af \u2192 \u23af\u23af\u23af\u23af \u23af \u2192 \u23af\u23af\u23af T 1 = 273 K T 2 = 273 K T 3 = 5 \u00d7 273 K V 1 = V V 2 = 5 V V 3 = 5 V P 1 = P 0 P P 2 5 = P 3 = P q total = nRT V V n C T T V m \u22c5 + \u22c5 \u22c5 \u2212 ln ( ) 2 1 3 2 1 or, 80 \u00d7 10 3 = 3 \u00d7 8.314 \u00d7 273 \u00d7 ln5 + 3 \u00d7 C V m , \u00d7 4 \u00d7 273 \u2234 C V m , \u2248 21 J/K-mol = 5 cal/K-mol
Answer: 3
\nOriginal PDF solution page
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Solution: PT = Constant \u21d2 P \u22c5 V 1/2 = Constant Now, C C R x f f m V m = + \u2212 \u21d2 = \u00d7 + \u2212 \u21d2 \u2248 1 1 29 8 314 2 8 314 1 1 2 3 . .
Answer: 9
\nOriginal PDF solution page
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Solution: u u T T 2 1 2 2 2 300 1200 = = \u21d2 = K \u21d2 T 2 = 1200 K Now, q n C T T V V m = \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 = , ( ) ( ) 2 1 56 28 5 2 2 1200 300 9000 cal
Answer: 6
\nOriginal PDF solution page
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Solution: Z u N RT M PN RT w aV A = \u22c5 \u22c5 = \u22c5 \u22c5 = 1 4 1 4 8 * \u03c0 Constant or, P T = \u21d2 \u22c5 = \u2212 Constant P V Constant 1 \u2234 C C R x R R R m V m = + \u2212 = + \u2212 \u2212 = = 1 1 5 2 1 1 3 6 ( ) cal/K mol
Answer: 5
\nOriginal PDF solution page
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Solution: \u0394 = \u22c5 \u0394 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 = \u2212 G V P 13 0 78 10 3001 1 10 5000 6 3 5 . ( ) m N m J 2
Answer: 8
\nOriginal PDF solution page
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Solution: \u0394 G 1 = \u0394 H \u2013 T 1 \u22c5 \u0394 S and \u0394 G 2 = \u0394 H \u2013 T 2 \u22c5 \u0394 S \u2234 (\u2013 \u0394 G 2 ) \u2013 (\u2013 \u0394 G 1 ) = ( T 2 \u2013 T 1 ) \u22c5 \u0394 S = ( T 2 \u2013 T 1 ) \u00d7 \u0394 \u2212 \u0394 H G T 1 1 = (302 \u2013 298) \u00d7 ( ) ( ) \u2212 \u2212 \u2212 = 5737 6333 298 8 kJ
Answer: 3
\nOriginal PDF solution page
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Solution: Graphite \u001f Diamond; \u0394 G \u00b0 = 5.0 kJ P = 1 bar \u0394 G = 0 P = ? Now, \u0394 G 2 \u2013 \u0394 G 1 = ( V P \u2013 V G ) ( P 2 \u2013 P 1 ) or, 0 5000 12 3 6 12 2 4 10 10 3 10 6 2 5 9 \u2212 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u00d7 \u2212 \u21d2 \u00d7 \u2212 . . ( ) P N m 2
Answer: 5
\nOriginal PDF solution page
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Solution: \u0394 G \u00b0 = \u2013 RT \u22c5 ln K eq = \u20132 \u00d7 300 \u00d7 ln ( e \u201310 ) = + 6000 cal Now, \u0394 \u00b0 = \u0394 \u00b0 \u2212 \u0394 \u00b0 = \u2212 = S H G T 7500 6000 300 5 cal/k-mol\n4.50 Chapter 4 HINTS AND EXPLANATIONS Four-digit Integer Type
Answer: 5713
\nOriginal PDF solution page
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Solution: w nRT V nb V nb an V V = \u2212 \u2212 \u2212 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ln 2 1 2 2 1 1 1 = \u2212 \u00d7 \u00d7 \u00d7 \u2212 \u00d7 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u00d7 \u00d7 \u00d7 \u2212 1 8 314 300 20 1 0 03 2 1 0 03 1 42 10 1 1 20 1 12 2 . ln . . . 2 2 10 10 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a5 \u2212 = \u20135713.16 J
Answer: 0044
\nOriginal PDF solution page
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Solution: \u0394 H = \u0394 U + \u0394 ( PV ) = 30 + (4 \u00d7 5 \u2013 2 \u00d7 3) = 44 L -atm
Answer: 0209
\nOriginal PDF solution page
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Solution: Calcite \u2192 Aragonite \u0394 H = \u0394 U + P \u22c5 \u0394 V \u0394 V = 210 J + 2 7 10 100 3 100 2 7 10 5 6 3 . . \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 N m m 2 = 209 J
Answer: 8400
\nOriginal PDF solution page
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Solution: \u0394 U = ( q + w ) path I = ( q + w ) path II or, 10 \u00d7 10 3 \u00d7 4.2 J + 0 = (11 \u00d7 10 3 \u00d7 4.2 J) + (\u20130.5 w max ) \u2234 w max = 8400 J
Answer: 0025
\nOriginal PDF solution page
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Solution: H = U + PV = 2.5 PV \u2234 \u0394 H = 2.5 \u00d7 10 200 100 10 25 5 3 3 N m m kJ 2 \u00d7 \u2212 \u00d7 = \u2212 ( ) 0 10 25 3 m kJ \u00d7 \u2212 \u00d7 = \u2212 ( )
Answer: 0020
\nOriginal PDF solution page
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Solution: q = \u0394 U \u2013 w = 1.5 nR \u22c5 \u0394 T + P ext \u22c5 A \u22c5 \u0394 l 42 = 1.5 \u00d7 1 \u00d7 8.314 \u00d7 2 + 100 \u00d7 10 3 \u00d7 8.5 \u00d7 \u00d7 10 -4 \u00d7 \u0394 l \u2234 \u0394 l \u2248 0.2 m
Answer: 6000
\nOriginal PDF solution page
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Solution: \u0394 H \u2013 \u0394 U = P ( V D \u2013 V G ) \u21d2 \u2212 = \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u00d7 \u2212 1000 12 3 6 12 2 4 10 6 P . . \u2234 P = 6000 \u00d7 10 5 Pa
Answer: 1567
\nOriginal PDF solution page
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Solution: Solid (70 C) Liquid (70 C) Liquid (450 C) Vapour (450 \u00b0 \u23af \u2192 \u23af \u00b0 \u23af \u2192 \u23af \u00b0 \u23af \u2192 \u23af 1 2 3 \u00b0 \u00b0 C) q = q 1 + q 2 + q 3 = 30 \u00d7 10 + 10 \u00d7 0.215 \u00d7 380 + 10 \u00d7 45 = 1567 cal
Answer: 0033
\nOriginal PDF solution page
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Solution: q = \u00d7 \u00d7 = 12 0 5 1805 6 1805 . J w = \u2013 P \u22c5 ( V g \u2013 V l ) = \u2013 P \u22c5 V g = \u2013 nRT = \u2212 \u00d7 \u00d7 = \u2212 0 9 18 8 314 373 155 . . J \u2234 \u0394 U = q + w = 1805 + (\u2013155) = 1650 J (for 0.9 g ) = \u00d7 \u00d7 = \u2212 1650 0 9 18 10 33 3 . kJ
Answer: 1990
\nOriginal PDF solution page
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Solution: q = \u0394 U \u2013 w = 0 \u21d2 \u0394 U = w = \u2013 P ext ( V 2 \u2013 V 1 ) = \u2013 100(\u20131) = 100 bar-ml Now, \u0394 H = \u0394 U + \u0394 PV = 100 + (100 \u00d7 99 \u2013 1 \u00d7 100) = 9900 bar-ml = 990 J
Answer: 1500
\nOriginal PDF solution page
Open page 235 in PDFExtracted text
Solution: q = \u0394 U \u2013 w = 0 \u21d2 \u0394 U = w = \u00d7 \u00d7 1 2 10 1000 bar-ml = 500 J V 990 1000 1001 bar 1 bar (ml) 2 1 P
Answer: 0332
\nOriginal PDF solution page
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Solution: For adiabatic process: T T V V 2 1 1 2 1 1 4 1 300 2 400 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 = \u2212 \u22c5 \u2212 \u03b3 ( ) K Now, w = w 1 + w 2 = \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u22c5 \u2212 nRT V V nC T T V m ln [ ( ) , 2 1 2 1 = [\u20131 \u00d7 8.3 \u00d7 300 \u00d7 ln 2] + 1 8 3 1 4 1 400 300 \u00d7 \u2212 \u00d7 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 . . ( ) = 332 J
Answer: 0500
\nOriginal PDF solution page
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Solution: q n C T V m = \u22c5 \u22c5 \u0394 , or, 50 \u00d7 166 \u00d7 t = 1 8 21 10 2 9 10 0 0821 290 5 8 3 2 20 3 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 . . . . \u2234 t = 500 sec
Answer: 1900
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Solution: \u0394 U = 3 \u00d7 1.5 R \u00d7 100 + 2 \u00d7 2.5 R \u00d7 100 = 1900 cal\n4.51 Thermodynamics HINTS AND EXPLANATIONS
Answer: 3120
\nOriginal PDF solution page
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Solution: V A (1200 K) (300 K) 64 atm 1 atm (300 K) B C P Path AB (Adiabatic): P P T T 2 1 1 2 1 1 1 1 3 5 1 64 1200 300 2 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u03b3 atm Now, w total = w AB + w BC = \u22c5 \u22c5 \u0394 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f n C T nRT P P V m B C , ln = \u00d7 \u00d7 \u2212 \u2212 \u00d7 \u00d7 \u00d7 1 3 2 300 1200 1 300 2 1 R R ( ) ln = \u20133120 cal
Answer: 0169
\nOriginal PDF solution page
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Solution: \u0394 \u00b0 = \u2212 + \u00d7 = \u2212 + \u00d7 S S S S CH grap H 4 ( ) . ( . . ) 2 186 2 6 0 2 130 6 2 = \u201381 J/K Now, \u0394 G \u00b0 = \u0394 H \u00b0 \u2013 T \u22c5 \u0394 S \u00b0 = \u2013 T \u22c5 \u0394 S univ or, (\u201375 \u00d7 10 3 ) \u2013 300 \u00d7 (\u201381) = \u2013300 \u00d7 \u0394 S univ \u21d2 \u0394 S univ = 169 J/K
Answer: 0300
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Solution: (\u2013 mgh ) \u00d7 N = (\u2013 \u0394 G \u00b0) or 50 \u00d7 10 \u00d7 2 \u00d7 N = 600 10 2 27 27 300 3 \u00d7 \u00d7 \u00d7 \u21d2 = N
Answer: 2870
\nOriginal PDF solution page
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Solution: (\u2013 \u0394 G ) = \u2013 ( \u0394 H \u2013 T \u22c5 \u0394 S ) = \u2212 \u2212 \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = ( ) 2808 310 200 1000 2870 kJ
Answer: 0012
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Answer: 2349
\nOriginal PDF solution page
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Solution: a = 2 (i, vi) b = 3 (ii, v, vii) c = 4 (iii, iv, viii, ix) d = 9 (all)