JEE Chemistry Solutions MCQs Set 08

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Chapter-wise Objective Questions: Solutions

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Multiple Choice Questions

Question 1: The molal elevation constant is the ratio of the elevation in boiling point to
(a) molarity
(b) molality
(c) mole fraction of solute
(d) mole fraction of solvent
Show Answer & Explanation

Answer: (b) molality

Explanation:
1. ΔTb = Kb m, so Kb = ΔTb/m.

Teacher's Note:
1. Kb is the elevation produced by a 1 molal solution.

Question 2: The elevation in boiling point was 0.52°C when 6 g of a compound X was dissolved in 100 g of water. The molecular weight of X is (K = 5.2 K mol−1 per 100 g H2O)
(a) 120
(b) 60
(c) 100
(d) 342
Show Answer & Explanation

Answer: (b) 60

Explanation:
1. Kb = 5.2 K per mole in 100 g, i.e. 0.52 K kg mol−1.
2. m = 0.52/0.52 = 1 molal, so 6 g in 0.1 kg is 0.1 mol.
3. M = 6/0.1 = 60 g mol−1.

Teacher's Note:
1. The constant is quoted for 100 g of solvent.

Question 3: An aqueous solution containing 1 g of urea boils at 100.25°C. The aqueous solution containing 3 g of glucose in the same volume will boil at
(a) 100.75°C
(b) 100.5°C
(c) 100°C
(d) 100.25°C
Show Answer & Explanation

Answer: (d) 100.25°C

Explanation:
1. Moles of urea = 1/60 = 0.0167; moles of glucose = 3/180 = 0.0167.
2. Equal moles in equal volume give equal ΔTb, so the boiling point is 100.25°C.

Teacher's Note:
1. Colligative properties depend on the number of particles.

Question 4: Pure benzene freezes at 5.45°C. A 0.374 molal solution of tetrachloroethane in benzene freezes at 3.55°C. The Kf for benzene is
(a) 5.08 K kg mol−1
(b) 508 K kg mol−1
(c) 0.508 K kg mol−1
(d) 50.8 K kg mol−1
Show Answer & Explanation

Answer: (a) 5.08 K kg mol−1

Explanation:
1. ΔTf = 5.45 − 3.55 = 1.90 K.
2. Kf = 1.90/0.374 = 5.08 K kg mol−1.

Teacher's Note:
1. Kf = ΔTf/m.

Question 5: An aqueous solution freezes at −0.186°C (Kf = 1.86 K kg mol−1, Kb = 0.512 K kg mol−1). The elevation in boiling point is
(a) 0.186
(b) 0.512
(c) \( \dfrac{0.512}{1.86} \)
(d) 0.0512
Show Answer & Explanation

Answer: (d) 0.0512

Explanation:
1. m = 0.186/1.86 = 0.1.
2. ΔTb = 0.512 × 0.1 = 0.0512 K.

Teacher's Note:
1. The same molality gives both ΔTf and ΔTb.

Question 6: The boiling point of an aqueous solution of a non-volatile solute is 100.15°C. What is the freezing point of the aqueous solution obtained by diluting it with an equal volume of water? (Kb = 0.512, Kf = 1.86 K molality−1)
(a) −0.544°C
(b) −0.512°C
(c) −0.272°C
(d) −1.86°C
Show Answer & Explanation

Answer: (c) −0.272°C

Explanation:
1. m = 0.15/0.512 = 0.293.
2. Dilution with equal volume halves it: m = 0.1465.
3. ΔTf = 1.86 × 0.1465 = 0.272, so freezing point = −0.272°C.

Teacher's Note:
1. Dilution reduces molality in proportion to the amount of solvent.

Question 7: Camphor is used as a solvent to determine the molecular weight of non-volatile solutes by the Rast method because for camphor
(a) the molal depression constant is high
(b) the melting point is high
(c) it is cheap
(d) all of these
Show Answer & Explanation

Answer: (a) the molal depression constant is high

Explanation:
1. A large Kf (about 40 K kg mol−1) gives a large measurable depression even for small amounts of solute.

Teacher's Note:
1. Rast method uses the large cryoscopic constant of camphor.

Question 8: In the depression of freezing point experiment, it is found that
(a) the vapour pressure of the solution is more than that of pure water
(b) the vapour pressure of the solution is less than that of pure solute
(c) only solute molecules solidify at the freezing point
(d) only solvent molecules solidify at the freezing point
Show Answer & Explanation

Answer: (d) only solvent molecules solidify at the freezing point

Explanation:
1. At the freezing point of a dilute solution only pure solvent separates as solid.
2. The solution has a lower vapour pressure than the pure solvent.

Teacher's Note:
1. Freezing point depression arises because the solution's vapour pressure is lowered.

Question 9: The molal depression constant of water is 1.86 K kg mol−1. 0.02 mol of urea dissolved in 100 g of water will produce a depression in freezing point of
(a) 0.186°C
(b) 0.372°C
(c) 1.86°C
(d) 3.72°C
Show Answer & Explanation

Answer: (b) 0.372°C

Explanation:
1. m = 0.02/0.1 = 0.2.
2. ΔTf = 1.86 × 0.2 = 0.372°C.

Teacher's Note:
1. Convert grams of solvent to kilograms.

Question 10: The freezing point of an aqueous solution containing 18 g of C6H12O6 in 1000 g of water is (Kf = 1.86 K molality−1)
(a) −0.186°C
(b) −0.372°C
(c) −0.54°C
(d) −0.72°C
Show Answer & Explanation

Answer: (a) −0.186°C

Explanation:
1. m = 0.1.
2. ΔTf = 0.186, so the freezing point is −0.186°C.

Teacher's Note:
1. Freezing point = 0 − ΔTf for water.

Question 11: A solution of 1.25 g of a non-electrolyte in 20 g of water freezes at 271.94 K. If Kf = 1.86 K molality−1, the molecular weight of the solute is
(a) 207.8 g mol−1
(b) 179.79 g mol−1
(c) 209.6 g mol−1
(d) 96.01 g mol−1
Show Answer & Explanation

Answer: (d) 96.01 g mol−1

Explanation:
1. ΔTf = 273.15 − 271.94 = 1.21 K.
2. M = Kf × w × 1000/(ΔTf × W) = 1.86 × 1.25 × 1000/(1.21 × 20) = 96.1 g mol−1.

Teacher's Note:
1. Use 273.15 K as the freezing point of pure water.

Question 12: 1.00 g of a non-electrolyte solute (molar mass 250 g mol−1) was dissolved in 51.2 g of benzene. If the freezing point depression constant Kf of benzene is 5.12 K kg mol−1, the freezing point of benzene will be lowered by
(a) 0.4 K
(b) 0.3 K
(c) 0.5 K
(d) 0.2 K
Show Answer & Explanation

Answer: (a) 0.4 K

Explanation:
1. m = (1/250)/0.0512 = 0.078125.
2. ΔTf = 5.12 × 0.078125 = 0.4 K.

Teacher's Note:
1. Moles of solute = mass/molar mass.

Question 13: A solution of sucrose (molar mass 342 g mol−1) is made by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution is (Kf for water = 1.86 K kg mol−1)
(a) −0.570°C
(b) −0.372°C
(c) −0.520°C
(d) +0.372°C
Show Answer & Explanation

Answer: (b) −0.372°C

Explanation:
1. m = 68.5/342 = 0.2.
2. ΔTf = 1.86 × 0.2 = 0.372, so freezing point = −0.372°C.

Teacher's Note:
1. Depression lowers the freezing point; the sign is negative.

Question 14: Boiling point of benzene is 353.23 K. When 1.8 g of a non-volatile solute is dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Given Kb (benzene) = 2.53 K kg mol−1, the molar mass of the solute is
(a) 58 g mol−1
(b) 120 g mol−1
(c) 116 g mol−1
(d) 60 g mol−1
Show Answer & Explanation

Answer: (a) 58 g mol−1

Explanation:
1. ΔTb = 354.11 − 353.23 = 0.88 K.
2. M = 2.53 × 1.8 × 1000/(0.88 × 90) = 57.5, about 58 g mol−1.

Teacher's Note:
1. M = Kb w × 1000/(ΔTb W).

Question 15: Addition of 1 kg of a non-volatile solute to a volatile solvent raises the boiling point from 350 K to 355 K and lowers the freezing point from 250 K to 220 K. If Kb is 15.5 K kg mol−1, then Kf will be
(a) 2.1 K kg mol−1
(b) 93.0 K kg mol−1
(c) 76 K kg mol−1
(d) 10 K kg mol−1
Show Answer & Explanation

Answer: (b) 93.0 K kg mol−1

Explanation:
1. ΔTb = 5 K and ΔTf = 30 K for the same molality m.
2. m = 5/15.5, so Kf = 30/m = 30 × 15.5/5 = 93.0.

Teacher's Note:
1. Kf/Kb = ΔTf/ΔTb for the same solution.

Question 16: 0.5 mol of NaCl is dissolved in 500 g of water (Kf = 1.8, Kb = 0.5 K kg mol−1). The freezing point and boiling point of the solution are
(a) −1.8°C, 100.5°C
(b) −3.6°C, 101°C
(c) −3.6°C, 99°C
(d) 0°C, 100°C
Show Answer & Explanation

Answer: (b) −3.6°C, 101°C

Explanation:
1. m = 0.5/0.5 = 1 molal; i = 2.
2. ΔTf = 2 × 1.8 × 1 = 3.6, so Tf = −3.6°C.
3. ΔTb = 2 × 0.5 × 1 = 1, so Tb = 101°C.

Teacher's Note:
1. Remember the van't Hoff factor for NaCl.

Question 17: 1 g of a solute is dissolved in 50 g of benzene and shows a depression in freezing point of 0.4 K. The molar mass of the solute is (Kf = 5.12 K kg mol−1)
(a) 512
(b) 256
(c) 728
(d) 738
Show Answer & Explanation

Answer: (b) 256

Explanation:
1. M = Kf × w × 1000/(ΔTf × W) = 5.12 × 1 × 1000/(0.4 × 50) = 256 g mol−1.

Teacher's Note:
1. Always use grams of solvent with the factor 1000.

Question 18: The depression in freezing point of water containing ethylene glycol is 7.4°C. The mass of ethylene glycol added to 500 g of water is (Kf = 1.86 K kg mol−1)
(a) 23.3 g
(b) 233.3 g
(c) 123.3 g
(d) 12.33 g
Show Answer & Explanation

Answer: (c) 123.3 g

Explanation:
1. m = 7.4/1.86 = 3.978 mol kg−1.
2. Moles in 0.5 kg = 1.989; mass = 1.989 × 62 = 123.3 g.

Teacher's Note:
1. Molar mass of C2H6O2 is 62.

Question 19: 45 g of a non-volatile solute A is dissolved in 600 g of water. If the depression in freezing point is 2.2 K, the molar mass of the solute is (Kf = 1.86 K kg mol−1)
(a) 126.8
(b) 31.7
(c) 63.4
(d) 158.5
Show Answer & Explanation

Answer: (c) 63.4

Explanation:
1. M = 1.86 × 45 × 1000/(2.2 × 600) = 63.4 g mol−1.

Teacher's Note:
1. Substitute directly into M = Kf w × 1000/(ΔTf W).

Question 20: Ethylene glycol is used as an antifreeze. The mass of ethylene glycol that should be added to 2 litres of water to prevent it from freezing at −24°C is (Kf = 1.86 K kg mol−1)
(a) 1600 g
(b) 1200 g
(c) 800 g
(d) 400 g
Show Answer & Explanation

Answer: (a) 1600 g

Explanation:
1. m = 24/1.86 = 12.9 mol kg−1.
2. Moles for 2 kg = 25.8; mass = 25.8 × 62 = 1600 g.

Teacher's Note:
1. Water 2 L is taken as 2 kg.

Question 21: At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52 K kg mol−1, the boiling point of this solution will be
(a) 101°C
(b) 100°C
(c) 102°C
(d) 103°C
Show Answer & Explanation

Answer: (a) 101°C

Explanation:
1. x(solute) = (760 − 732)/760 = 0.0368.
2. n/(n + 5.556) = 0.0368 gives n = 0.2125 mol, so m = 2.125 mol kg−1.
3. ΔTb = 0.52 × 2.125 = 1.1, so the boiling point is about 101°C.

Teacher's Note:
1. Water boils at 100°C when P = 760 mm.

Question 22: Which of the following statements is correct for a solution in a freezing point depression experiment?
(a) Solution freezes at a temperature higher than the pure solvent
(b) Solution freezes at a lower temperature than the pure solvent
(c) Solution freezes at the same temperature as the pure solvent
(d) The freezing temperature is independent of the solute amount
Show Answer & Explanation

Answer: (b) Solution freezes at a lower temperature than the pure solvent

Explanation:
1. The vapour pressure curve of the solution lies below that of the solvent, so it meets the solid curve at a lower temperature.

Teacher's Note:
1. ΔTf = Kf m increases with concentration.

Question 23: Which of the following is not a colligative property?
(a) Osmotic pressure
(b) Relative lowering in vapour pressure
(c) Depression in freezing point
(d) Refractive index
Show Answer & Explanation

Answer: (d) Refractive index

Explanation:
1. Colligative properties depend on the number of solute particles; refractive index depends on the nature of the solute.

Teacher's Note:
1. The four colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.

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