JEE Chemistry Solutions MCQs Set 10

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Multiple Choice Questions

Question 1: Equimolar solutions of A and B show depression in freezing point in the ratio 2 : 1. A remains in the normal state in solution. B will be in
(a) normal state
(b) associated state
(c) hydrolysed state
(d) dissociated state
Show Answer & Explanation

Answer: (b) associated state

Explanation:
1. ΔTf ∝ i. A has i = 1, so B has i = 0.5.
2. i < 1 means B associates in solution.

Teacher's Note:
1. i > 1 means dissociation; i < 1 means association.

Question 2: The van't Hoff factor is
(a) less than one in case of dissociation
(b) more than one in case of association
(c) always less than one
(d) less than one in case of association
Show Answer & Explanation

Answer: (d) less than one in case of association

Explanation:
1. Association reduces the number of particles, so i < 1.
2. Dissociation gives i > 1.

Teacher's Note:
1. i = observed colligative property / calculated value.

Question 3: The van't Hoff factor (i) for a dilute solution of K3[Fe(CN)6] is
(a) 10
(b) 4
(c) 5
(d) 0.25
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3−: 4 ions.
2. i = 4 for complete dissociation.

Teacher's Note:
1. Complex ions do not dissociate further.

Question 4: The experimental molecular weight of an electrolyte will always be less than its calculated value because the value of the van't Hoff factor i is
(a) less than 1
(b) greater than 1
(c) one
(d) zero
Show Answer & Explanation

Answer: (b) greater than 1

Explanation:
1. Mobserved = Mcalc/i.
2. For electrolytes i > 1, so the observed value is smaller.

Teacher's Note:
1. Dissociation increases the number of particles.

Question 5: The van't Hoff factor (i) for a dilute aqueous solution of glucose is
(a) zero
(b) 1.0
(c) 1.5
(d) 2.0
Show Answer & Explanation

Answer: (b) 1.0

Explanation:
1. Glucose is a non-electrolyte and neither dissociates nor associates, so i = 1.

Teacher's Note:
1. Only electrolytes show i > 1.

Question 6: A 0.585% NaCl solution at 27°C has osmotic pressure (R = 0.0821 L atm K−1 mol−1)
(a) 2.49 atm
(b) 5 atm
(c) 1.2 atm
(d) 3.8 atm
Show Answer & Explanation

Answer: (b) 5 atm

Explanation:
1. 0.585% w/V = 5.85 g L−1 = 0.1 M.
2. π = iCRT = 2 × 0.1 × 0.0821 × 300 = 4.93, about 5 atm.

Teacher's Note:
1. Include i = 2 for NaCl.

Question 7: A substance A when dissolved in solvent B shows the molecular mass corresponding to A3. The van't Hoff factor will be
(a) 1
(b) 2
(c) 3
(d) \( \dfrac13 \)
Show Answer & Explanation

Answer: (d) \( \dfrac13 \)

Explanation:
1. Three molecules associate into one: observed particles = 1/3 of expected.
2. i = 1/3.

Teacher's Note:
1. Association: i < 1.

Question 8: The ratio of the value of any colligative property for KCl solution to that for sugar solution (same concentration) is nearly
(a) 1
(b) 0.5
(c) 2
(d) 2.5
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. KCl gives 2 ions (i = 2) and sugar gives 1 particle (i = 1).
2. Ratio = 2.

Teacher's Note:
1. The ratio equals the ratio of van't Hoff factors.

Question 9: The lowering of vapour pressure of 0.1 M aqueous solutions of NaCl, CuSO4 and K2SO4 are in the ratio
(a) all equal
(b) 1 : 1 : 1.5
(c) 3 : 2 : 1
(d) 1.5 : 1 : 2.5
Show Answer & Explanation

Answer: (b) 1 : 1 : 1.5

Explanation:
1. i values: NaCl 2, CuSO4 2, K2SO4 3.
2. Ratio 2 : 2 : 3 = 1 : 1 : 1.5.

Teacher's Note:
1. Lowering of vapour pressure ∝ i × concentration.

Question 10: The freezing point of a 1 molal NaCl solution, assuming NaCl to be 100% dissociated in water, is (Kf = 1.86 K molality−1)
(a) −1.86°C
(b) −3.72°C
(c) +1.86°C
(d) +3.72°C
Show Answer & Explanation

Answer: (b) −3.72°C

Explanation:
1. ΔTf = i Kf m = 2 × 1.86 × 1 = 3.72, so the freezing point is −3.72°C.

Teacher's Note:
1. Freezing point is depressed, so negative.

Question 11: The molal elevation constant of water is 0.51. The boiling point of a 0.1 molal aqueous NaCl solution is nearly
(a) 100.05°C
(b) 100.1°C
(c) 100.2°C
(d) 101.0°C
Show Answer & Explanation

Answer: (b) 100.1°C

Explanation:
1. ΔTb = 2 × 0.51 × 0.1 = 0.102.
2. Boiling point = 100.1°C.

Teacher's Note:
1. Boiling point is raised by i Kb m.

Question 12: The molal elevation constant of water is 0.52 K molality−1. The boiling point of a 1.0 molal aqueous KCl solution (assuming complete dissociation) should be
(a) 100.52°C
(b) 101.04°C
(c) 99.48°C
(d) 98.96°C
Show Answer & Explanation

Answer: (b) 101.04°C

Explanation:
1. ΔTb = 2 × 0.52 × 1 = 1.04.
2. Boiling point = 101.04°C.

Teacher's Note:
1. Boiling point is raised, not lowered.

Question 13: Solute A is a ternary electrolyte and solute B is a non-electrolyte. If a 0.1 M solution of B produces an osmotic pressure of 2P, then a 0.05 M solution of A at the same temperature (assuming complete dissociation) will produce an osmotic pressure equal to
(a) P
(b) 1.5P
(c) 2P
(d) 3P
Show Answer & Explanation

Answer: (d) 3P

Explanation:
1. For B: 0.1 RT = 2P, so RT = 20P.
2. For A, i = 3: π = 3 × 0.05 × 20P = 3P.

Teacher's Note:
1. A ternary electrolyte gives 3 ions.

Question 14: Phenol associates in benzene as C6H5OH ⇌ 1/2 (C6H5OH)2. If the degree of association of phenol is 40%, the van't Hoff factor i is
(a) 1
(b) 0.8
(c) 1.4
(d) 0.6
Show Answer & Explanation

Answer: (b) 0.8

Explanation:
1. i = 1 − α + α/2 = 1 − α/2 = 1 − 0.2 = 0.8.

Teacher's Note:
1. Dimerisation leaves 1/2 particle per molecule associated.

Question 15: The observed and calculated molecular weights of calcium nitrate are 65.6 and 164 respectively. The degree of dissociation of calcium nitrate is
(a) 25%
(b) 50%
(c) 75%
(d) 60%
Show Answer & Explanation

Answer: (c) 75%

Explanation:
1. i = 164/65.6 = 2.5.
2. Ca(NO3)2 gives 3 ions: i = 1 + 2α, so α = 0.75 = 75%.

Teacher's Note:
1. i = Mcalc/Mobs.

Question 16: A 0.004 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at 25°C. The apparent degree of dissociation of Na2SO4 is
(a) 25%
(b) 50%
(c) 75%
(d) 85%
Show Answer & Explanation

Answer: (c) 75%

Explanation:
1. Isotonic: i × 0.004 = 0.010, so i = 2.5.
2. i = 1 + 2α gives α = 0.75 = 75%.

Teacher's Note:
1. Na2SO4 gives 3 ions.

Question 17: A 5.8% (w/V) NaCl solution will exert an osmotic pressure closest to which one of the following?
(a) 5.8% (w/V) sucrose solution
(b) 5.8% (w/V) glucose solution
(c) 2 M sucrose solution
(d) 1 M glucose solution
Show Answer & Explanation

Answer: (c) 2 M sucrose solution

Explanation:
1. 5.8% NaCl = 58 g L−1 ≈ 1 M; with i = 2 the effective concentration is about 2 M.
2. It matches a 2 M non-electrolyte solution.

Teacher's Note:
1. Compare the total particle concentration.

Question 18: 0.5 molal aqueous solution of a weak acid HX is 20% ionised. If Kf for water is 1.86 K kg mol−1, the lowering in freezing point of the solution is
(a) −0.56 K
(b) −1.12 K
(c) 0.56 K
(d) 1.12 K
Show Answer & Explanation

Answer: (d) 1.12 K

Explanation:
1. i = 1 + α = 1.2.
2. ΔTf = 1.2 × 1.86 × 0.5 = 1.116 K, about 1.12 K (a lowering, quoted as positive).

Teacher's Note:
1. The lowering is a positive magnitude; the freezing point itself falls.

Question 19: A 0.0020 m aqueous solution of an ionic compound [Co(NH3)5(NO2)]Cl freezes at −0.00732°C. The number of moles of ions which 1 mol of the ionic compound produces on being dissolved in water will be (Kf = 1.86°C m−1)
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. i = ΔTf/(Kf m) = 0.00732/(1.86 × 0.0020) = 1.97, about 2.
2. It gives [Co(NH3)5NO2]+ and Cl−.

Teacher's Note:
1. The complex cation stays intact.

Question 20: The freezing point depression constant for water is 1.86°C m−1. If 5 g of Na2SO4 is dissolved in 45.0 g of water, the freezing point is changed by −3.82°C. The van't Hoff factor for Na2SO4 is (Na = 23, S = 32, O = 16)
(a) 2.05
(b) 2.63
(c) 3.11
(d) 0.381
Show Answer & Explanation

Answer: (b) 2.63

Explanation:
1. m = (5/142)/0.045 = 0.7825.
2. i = 3.82/(1.86 × 0.7825) = 2.62.

Teacher's Note:
1. Experimental i is below the ideal value 3.

Question 21: The van't Hoff factor i for a compound which undergoes dissociation in one solvent and association in another solvent is respectively
(a) less than one and greater than one
(b) less than one and less than one
(c) greater than one and less than one
(d) greater than one and greater than one
Show Answer & Explanation

Answer: (c) greater than one and less than one

Explanation:
1. Dissociation increases particles (i > 1) and association decreases them (i < 1).

Teacher's Note:
1. Order matters: dissociation first, association second.

Question 22: The boiling point of a 0.2 mol kg−1 solution of X in water is greater than that of an equimolal solution of Y in water. Which statement is true?
(a) The molecular mass of X is greater than that of Y
(b) The molecular mass of X is less than that of Y
(c) Y is undergoing dissociation in water while X undergoes no change
(d) X is undergoing dissociation in water while Y undergoes no change
Show Answer & Explanation

Answer: (d) X is undergoing dissociation in water while Y undergoes no change

Explanation:
1. At equal molality, a higher boiling point means more particles per formula unit (greater i).
2. So X dissociates while Y stays unchanged.

Teacher's Note:
1. Molecular mass does not matter at equal molality.

Question 23: Which one of the following electrolytes has the same value of van't Hoff factor i as that of Al2(SO4)3 (if all are 100% ionised)?
(a) K3[Fe(CN)6]
(b) Al(NO3)3
(c) K4[Fe(CN)6]
(d) K2SO4
Show Answer & Explanation

Answer: (c) K4[Fe(CN)6]

Explanation:
1. Al2(SO4)3 → 2Al3+ + 3SO42−: i = 5.
2. K4[Fe(CN)6] → 4K+ + [Fe(CN)6]4−: i = 5.

Teacher's Note:
1. K3[Fe(CN)6] has 4, Al(NO3)3 has 4 and K2SO4 has 3.

Question 24: The van't Hoff factor i for a dilute aqueous solution of the strong electrolyte barium hydroxide is
(a) 2
(b) 3
(c) 0
(d) 1
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Ba(OH)2 → Ba2+ + 2OH−: 3 ions, i = 3.

Teacher's Note:
1. Strong electrolyte: complete dissociation in dilute solution.

Question 25: 0.04 molal aqueous solution of a monoprotic acid has a depression in freezing point equal to 0.1°C. The van't Hoff factor of the acid is (Kf = 1.86 K kg mol−1)
(a) 1
(b) 2
(c) 2.5
(d) 1.34
Show Answer & Explanation

Answer: (d) 1.34

Explanation:
1. i = ΔTf/(Kf m) = 0.1/(1.86 × 0.04) = 1.34.

Teacher's Note:
1. i < 2 shows the acid is only partly ionised (α = 0.34).

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