JEE Chemistry Solutions MCQs Set 05

Chemistry Objective Questions and Answers: Solutions

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Multiple Choice Questions

Question 1: At 25°C the vapour pressure of pure methyl alcohol is 96.0 torr. What is the mole fraction of CH\(_3\)OH in a solution in which the partial vapour pressure of CH\(_3\)OH is 23.0 torr at 25°C?
(a) 0.76
(b) 0.12
(c) 0.24
(d) 4.17
Show Answer & Explanation

Answer: (c) 0.24

Explanation:
1. By Raoult's law, \( p = x\,p^0 \).
2. \( x = 23.0/96.0 = 0.24 \).

Teacher's Note:
1. The partial pressure of a component equals its mole fraction times its pure vapour pressure.
2. The ratio must always be less than 1 for a real solution.

Question 2: The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added, the vapour pressure drops to 0.60 atm. The mole fraction of B in the solution is
(a) 0.25
(b) 0.75
(c) 0.20
(d) 0.40
Show Answer & Explanation

Answer: (a) 0.25

Explanation:
1. Relative lowering of vapour pressure equals the mole fraction of the solute.
2. \( x_B = (0.80-0.60)/0.80 = 0.25 \).

Teacher's Note:
1. Relative lowering of vapour pressure is a colligative property equal to the mole fraction of the non-volatile solute.
2. It does not depend on the nature of the solute.

Question 3: The vapour pressure of pure water at 26°C is 25.21 torr. What is the vapour pressure of a solution containing 20.0 g of glucose (C\(_6\)H\(_{12}\)O\(_6\)) in 70 g of water?
(a) 25.2 torr
(b) 23.4 torr
(c) 0.99 torr
(d) 24.5 torr
Show Answer & Explanation

Answer: (d) 24.5 torr

Explanation:
1. Moles of glucose = 20/180 = 0.111; moles of water = 70/18 = 3.889.
2. \( x_{water} = 3.889/(3.889+0.111) = 0.9722 \).
3. \( p = 0.9722 \times 25.21 = 24.5 \) torr.

Teacher's Note:
1. Use the mole fraction of the solvent when finding the vapour pressure of the solution.
2. Check that the answer is slightly less than the pure solvent value.

Question 4: The vapour pressure of pure water at 25°C is 23.76 torr. A solution of 5.40 g of a non-volatile substance in 90.0 g of water has a vapour pressure of 23.32 torr. The molecular weight of the solute is
(a) 5.7 g/mol
(b) 57.24 g/mol
(c) 114.5 g/mol
(d) 28.6 g/mol
Show Answer & Explanation

Answer: (b) 57.24 g/mol

Explanation:
1. \( (p^0-p)/p^0 = 0.44/23.76 = 0.01852 \).
2. \( n_2/(n_1+n_2)=0.01852 \) with \( n_1 = 90/18 = 5 \), so \( n_2 = 0.0944 \) mol.
3. \( M = 5.40/0.0943 = 57.2 \) g/mol.

Teacher's Note:
1. Convert relative lowering to moles of solute using the exact mole fraction relation.
2. Do not use the approximation n2/n1 when the solution is not very dilute.

Question 5: The vapour pressures of ethanol and methanol are 44.5 mm and 88.7 mm Hg. An ideal solution is made by mixing 60 g of ethanol with 40 g of methanol. The total vapour pressure is
(a) 133.2 mm Hg
(b) 66.1 mm Hg
(c) 44.5 mm Hg
(d) 77.3 mm Hg
Show Answer & Explanation

Answer: (b) 66.1 mm Hg

Explanation:
1. Moles ethanol = 60/46 = 1.304; moles methanol = 40/32 = 1.25.
2. \( x_e = 0.5105,\ x_m = 0.4895 \).
3. \( P = 0.5105(44.5)+0.4895(88.7) = 66.1 \) mm Hg.

Teacher's Note:
1. For an ideal solution the total pressure is the mole-fraction weighted sum of the pure vapour pressures.
2. The result lies between the two pure values.

Question 6: A benzene-toluene solution has mole fraction of toluene 0.50. Vapour pressures of pure benzene and toluene are 119 and 37 torr. The mole fraction of toluene in the vapour in equilibrium is
(a) 0.50
(b) 0.763
(c) 0.31
(d) 0.237
Show Answer & Explanation

Answer: (d) 0.237

Explanation:
1. \( p_t = 0.5\times37 = 18.5 \) torr and \( p_b = 0.5\times119 = 59.5 \) torr.
2. \( y_t = 18.5/78 = 0.237 \).

Teacher's Note:
1. The vapour is richer in the more volatile component, benzene.
2. Mole fraction in vapour equals partial pressure divided by total pressure (Dalton's law).

Question 7: A benzene-toluene solution at 30°C has mole fraction of benzene 0.60. With \( P^0_b = 119 \) torr and \( P^0_t = 37 \) torr, the mole fraction of benzene in the vapour is
(a) 0.829
(b) 0.60
(c) 0.682
(d) 0.171
Show Answer & Explanation

Answer: (a) 0.829

Explanation:
1. \( p_b = 0.6\times119 = 71.4 \), \( p_t = 0.4\times37 = 14.8 \); total 86.2 torr.
2. \( y_b = 71.4/86.2 = 0.829 \).

Teacher's Note:
1. Combine Raoult's law with Dalton's law to get vapour composition.
2. Compare with the 0.682 value obtained for liquid composition 0.40 to see the trend.

Question 8: At 90°C the vapour pressure of toluene is 400 torr and of o-xylene is 150 torr. A liquid mixture boils at 90°C when the pressure is 0.50 atm. The mole fraction of toluene in the liquid mixture is
(a) 0.08
(b) 0.50
(c) 0.92
(d) 0.97
Show Answer & Explanation

Answer: (c) 0.92

Explanation:
1. Boiling means total pressure = 380 torr (0.5 atm).
2. \( x_t\cdot400+(1-x_t)150=380 \), so \( 250x_t = 230 \) and \( x_t = 0.92 \).

Teacher's Note:
1. A liquid boils when its total vapour pressure equals the external pressure.
2. Convert atm to torr before solving.

Question 9: Two liquids A and B form an ideal solution. At a temperature T the total vapour pressure is 400 torr, the mole fraction of A in the vapour is 0.40 and in the liquid 0.75. The vapour pressures of pure A and pure B are respectively
(a) 533 torr and 640 torr
(b) 300 torr and 500 torr
(c) 160 torr and 240 torr
(d) 213.3 torr and 960 torr
Show Answer & Explanation

Answer: (d) 213.3 torr and 960 torr

Explanation:
1. \( p_A = 0.40\times400 = 160 \) torr and \( p_B = 240 \) torr.
2. \( P^0_A = 160/0.75 = 213.3 \) torr; \( P^0_B = 240/0.25 = 960 \) torr.

Teacher's Note:
1. Partial pressure = mole fraction in vapour times total pressure, and also = mole fraction in liquid times pure vapour pressure.
2. Equate the two to get the pure vapour pressures.

Question 10: The relative lowering in vapour pressure when 100 g of a non-volatile solute (molar mass 100) is dissolved in 432 g of water is
(a) 0.04
(b) 0.96
(c) 0.25
(d) 0.40
Show Answer & Explanation

Answer: (a) 0.04

Explanation:
1. Moles of solute = 1; moles of water = 432/18 = 24.
2. \( x_{solute} = 1/25 = 0.04 \).

Teacher's Note:
1. Relative lowering equals the mole fraction of the solute.
2. Always compute moles first.

Question 11: What mass of urea must be dissolved in 100 g of water to decrease the vapour pressure of water by 25%?
(a) 33.3 g
(b) 55.5 g
(c) 111.1 g
(d) 25 g
Show Answer & Explanation

Answer: (c) 111.1 g

Explanation:
1. Mole fraction of urea must be 0.25, so \( n_u/(n_u+5.556)=0.25 \).
2. \( n_u = 1.852 \) mol; mass = 1.852 × 60 = 111.1 g.

Teacher's Note:
1. A 25% decrease means the relative lowering is 0.25.
2. Use the mole fraction of the solute, not of the solvent.

Question 12: The vapour pressure of an aqueous glucose solution is 750 mm Hg at 373 K. The molality of the solution is
(a) 1.35 m
(b) 0.741 m
(c) 0.013 m
(d) 7.41 m
Show Answer & Explanation

Answer: (b) 0.741 m

Explanation:
1. At 373 K pure water has 760 mm Hg, so \( x_{glucose} = 10/760 = 0.01316 \).
2. \( n_g/n_w = 0.01316/0.98684 \); molality = 0.01333 × 55.5 = 0.741 m.

Teacher's Note:
1. Pure water has a vapour pressure of 760 mm Hg at its boiling point.
2. Molality is moles per 1000 g of water.

Question 13: The vapour pressure of pure benzene at 25°C is 639.7 mm Hg and that of a solution of a solute in benzene is 631.7 mm Hg. The molality of the solution is
(a) 0.162 m
(b) 1.62 m
(c) 0.0125 m
(d) 0.0162 m
Show Answer & Explanation

Answer: (a) 0.162 m

Explanation:
1. \( x_{solute} = 8.0/639.7 = 0.0125 \).
2. \( n_2/n_1 = 0.01266 \); with \( n_1 = 1000/78 = 12.82 \), molality = 0.162 m.

Teacher's Note:
1. Convert mole fraction to molality using moles of solvent per kilogram.
2. Benzene has a molar mass of 78 g/mol.

Question 14: The vapour pressure of pure benzene is 640 mm Hg. A non-volatile non-electrolyte (2.175 g) dissolved in 39.0 g of benzene gives a vapour pressure of 600 mm Hg. The molar mass of the solid is
(a) 32.6 g/mol
(b) 130.5 g/mol
(c) 65.25 g/mol
(d) 6.5 g/mol
Show Answer & Explanation

Answer: (c) 65.25 g/mol

Explanation:
1. \( x_{solute} = 40/640 = 0.0625 \), so \( n_2/n_1 = 0.0667 \).
2. \( n_1 = 39/78 = 0.5 \), so \( n_2 = 0.0333 \) mol.
3. \( M = 2.175/0.0333 = 65.25 \).

Teacher's Note:
1. Be careful: the relative lowering gives mole fraction, not the ratio n2/n1.
2. Convert it before using the benzene moles.

Question 15: The vapour pressure of water is 17.54 mm Hg at 293 K. The vapour pressure of a 0.5 molal solution of a solute in it is
(a) 8.77 mm Hg
(b) 17.38 mm Hg
(c) 17.54 mm Hg
(d) 17.7 mm Hg
Show Answer & Explanation

Answer: (b) 17.38 mm Hg

Explanation:
1. \( n_1 = 1000/18 = 55.56 \); \( x_2 = 0.5/56.06 = 0.00892 \).
2. \( p = 17.54(1-0.00892) = 17.38 \) mm Hg.

Teacher's Note:
1. Molality gives moles of solute per 55.56 moles of water.
2. The vapour pressure of the solution is always lower than that of the pure solvent.

Question 16: Benzene and toluene form ideal solutions A and B at 313 K. Solution A contains equal moles of benzene and toluene (total pressure \( P_A \)); solution B contains equal masses of both (total pressure \( P_B \)). Vapour pressures of pure benzene and toluene are 160 and 60 mm Hg. The value of \( P_A/P_B \) is
(a) 1.04
(b) 0.5
(c) 1.5
(d) 0.964
Show Answer & Explanation

Answer: (d) 0.964

Explanation:
1. \( P_A = 0.5(160)+0.5(60) = 110 \) mm Hg.
2. Equal masses: moles benzene 1/78 and toluene 1/92 give \( x_b = 0.5413 \).
3. \( P_B = 0.5413(160)+0.4587(60) = 114.1 \); ratio = 0.964.

Teacher's Note:
1. Equal masses do not mean equal moles.
2. Compute mole fractions in each case before using Raoult's law.

Question 17: When 10.6 g of a non-volatile substance is dissolved in 740 g of ether, its boiling point rises by 0.284°C. The molar mass of the substance is (\( K_b \) of ether = 2.11 °C kg/mol)
(a) 53 g/mol
(b) 212 g/mol
(c) 106 g/mol
(d) 10.6 g/mol
Show Answer & Explanation

Answer: (c) 106 g/mol

Explanation:
1. \( \Delta T_b = K_b\,\dfrac{w\times1000}{M\,W} \).
2. \( M = 2.11\times10.6\times1000/(0.284\times740) = 106 \) g/mol.

Teacher's Note:
1. Use the elevation formula with the solvent mass in grams.
2. Check the units of Kb.

Question 18: A solution containing 3.24 g of a non-volatile non-electrolyte in 200 g of water boils at 100.130°C at 1 atm. The molar mass of the solute is (\( K_b \) of water = 0.513°C/m)
(a) 64.0 g/mol
(b) 32.0 g/mol
(c) 128 g/mol
(d) 6.4 g/mol
Show Answer & Explanation

Answer: (a) 64.0 g/mol

Explanation:
1. \( m = 0.130/0.513 = 0.2534 \) mol/kg.
2. Moles = 0.2534 × 0.2 = 0.05068; \( M = 3.24/0.05068 = 63.9 \) g/mol.

Teacher's Note:
1. Find molality from the boiling-point elevation first.
2. Then multiply by the solvent mass in kg to get moles.

Question 19: The molar mass of an organic compound is 58.0 g/mol. The boiling point of a solution containing 24.0 g of it in 600 g of water, when pure water boils at 99.725°C at the existing pressure, is (\( K_b = 0.512 \))
(a) 99.804°C
(b) 100.512°C
(c) 99.646°C
(d) 100.079°C
Show Answer & Explanation

Answer: (d) 100.079°C

Explanation:
1. Molality = (24/58)/0.6 = 0.690 m.
2. \( \Delta T_b = 0.512\times0.690 = 0.353 \).
3. Boiling point = 99.725 + 0.353 = 100.078, that is about 100.079°C.

Teacher's Note:
1. Add the elevation to the actual boiling point of the pure solvent under the prevailing pressure, not to 100°C.
2. The solvent boiling point here is given as 99.725°C.

Question 20: An aqueous solution of a non-volatile solute boils at 100.17°C. It will freeze at (\( K_b = 0.512,\ K_f = 1.86 \) for water)
(a) -0.17°C
(b) -0.62°C
(c) -1.86°C
(d) 0.62°C
Show Answer & Explanation

Answer: (b) -0.62°C

Explanation:
1. \( m = 0.17/0.512 = 0.332 \).
2. \( \Delta T_f = 1.86\times0.332 = 0.618 \), so the freezing point is -0.62°C.

Teacher's Note:
1. The same molality governs both the boiling-point elevation and the freezing-point depression.
2. Water freezes at 0°C.

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