JEE Chemistry Solutions MCQs Set 28

Multiple Choice Questions (MCQs) for JEE Chemistry: Solutions

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Practice Solutions MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: An automobile radiator holds 1.0 kg of water. How many grams of ethylene glycol must be added to lower the freezing point to -2.8 C? (\( K_f=1.86 \) K kg mol\(^{-1}\), M = 62)
(a) 72 g
(b) 39 g
(c) 93 g
(d) 27 g
Show Answer & Explanation

Answer: (c) 93 g

Explanation:
1. m = 2.8/1.86 = 1.505 mol/kg.
2. Mass = 1.505 x 62 = 93.3 g.

Teacher's Note:
1. The solution has 1 kg of water.
2. Ethylene glycol is a non-electrolyte.

Question 2: The vapour pressure of acetone at 20 C is 185 torr. When 1.2 g of a non-volatile substance is dissolved in 100 g of acetone, the vapour pressure becomes 183 torr. The molar mass of the substance is:
(a) 64 g/mol
(b) 32 g/mol
(c) 128 g/mol
(d) 488 g/mol
Show Answer & Explanation

Answer: (a) 64 g/mol

Explanation:
1. \( X_2=\dfrac{185-183}{185}=0.01081 \); moles of acetone = 100/58 = 1.724.
2. \( n_2=\dfrac{X_2n_1}{1-X_2}=0.01884 \) mol.
3. M = 1.2/0.01884 = 63.7, about 64 g/mol.

Teacher's Note:
1. Acetone M = 58.
2. Using the dilute approximation gives 64 g/mol as well.

Question 3: 18 g of glucose is added to 178.2 g of water. The vapour pressure of water (in torr) for this solution is (vapour pressure of pure water at this temperature 760 torr):
(a) 76.0
(b) 759.0
(c) 7.6
(d) 752.4
Show Answer & Explanation

Answer: (d) 752.4

Explanation:
1. Moles: glucose 0.1; water 9.9.
2. \( X_w=0.99 \); \( p=0.99\times760=752.4 \) torr.

Teacher's Note:
1. Relative lowering = 0.01.
2. Lowering = 7.6 torr.

Question 4: The freezing point of benzene decreases by 0.45 C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, the percentage association is: (\( K_f=5.12 \) K kg mol\(^{-1}\))
(a) 80.4%
(b) 74.6%
(c) 94.6%
(d) 64.6%
Show Answer & Explanation

Answer: (c) 94.6%

Explanation:
1. m = (0.2/60)/0.020 = 0.1667 mol/kg.
2. \( i=\dfrac{0.45}{5.12\times0.1667}=0.527 \).
3. For dimerisation \( i=1-\alpha/2 \), so \( \alpha=0.946 \).

Teacher's Note:
1. 2 CH3COOH gives (CH3COOH)2.
2. i < 1 shows association.

Question 5: For 1 molal aqueous solutions of the following compounds, which shows the highest freezing point?
(a) \( [\ce{Co(H2O)3Cl3}]\ce{.3H2O} \)
(b) \( [\ce{Co(H2O)4Cl2}]\ce{Cl.2H2O} \)
(c) \( [\ce{Co(H2O)6}]\ce{Cl3} \)
(d) \( [\ce{Co(H2O)5Cl}]\ce{Cl2.H2O} \)
Show Answer & Explanation

Answer: (a) \( [\ce{Co(H2O)3Cl3}]\ce{.3H2O} \)

Explanation:
1. Count the particles furnished in water: [Co(H2O)3Cl3] is a neutral complex (all three chlorides are inside the sphere), so i = 1.
2. [Co(H2O)4Cl2]Cl gives 2 ions, [Co(H2O)5Cl]Cl2 gives 3 and [Co(H2O)6]Cl3 gives 4.
3. The fewest particles give the smallest depression and hence the highest freezing point.

Teacher's Note:
1. Chloride ions inside the coordination sphere do not ionise.
2. Water of crystallisation does not add particles.

Question 6: A solution at 20 C contains 1.5 mol benzene and 3.5 mol toluene (\( p^0 \): benzene 74.7 torr, toluene 22.3 torr). The total vapour pressure and the mole fraction of benzene in the vapour are, respectively:
(a) 35.0 torr and 0.480
(b) 30.5 torr and 0.389
(c) 35.8 torr and 0.280
(d) 38.0 torr and 0.589
Show Answer & Explanation

Answer: (d) 38.0 torr and 0.589

Explanation:
1. \( X_B=1.5/5=0.3 \); \( p_B=0.3\times74.7=22.41 \); \( p_T=0.7\times22.3=15.61 \).
2. Total = 38.02 torr; \( y_B=22.41/38.02=0.589 \).

Teacher's Note:
1. The vapour is richer in benzene than the liquid (0.3).
2. Dalton's law gives the vapour composition.

Question 7: The solubility of \( \ce{N2} \) in water at 300 K and 500 torr partial pressure is 0.01 g/L. The solubility at 750 torr partial pressure is:
(a) 0.02 g/L
(b) 0.015 g/L
(c) 0.0075 g/L
(d) 0.005 g/L
Show Answer & Explanation

Answer: (b) 0.015 g/L

Explanation:
1. Henry's law: solubility is proportional to partial pressure.
2. \( S=0.01\times750/500=0.015 \) g/L.

Teacher's Note:
1. Solubility increases with pressure.
2. Temperature is constant.

Question 8: 5 g of \( \ce{Na2SO4} \) (M = 142) is dissolved in x g of water. The freezing point change is 3.82 C. If \( \ce{Na2SO4} \) is 81.5% ionised (\( K_f=1.86 \)), the value of x is approximately:
(a) 65 g
(b) 25 g
(c) 45 g
(d) 15 g
Show Answer & Explanation

Answer: (c) 45 g

Explanation:
1. \( i=1+2(0.815)=2.63 \); moles = 5/142 = 0.0352.
2. \( 3.82=2.63\times1.86\times m \), so m = 0.781 mol/kg.
3. Mass of water = 0.0352/0.781 = 0.0451 kg = 45 g.

Teacher's Note:
1. Na2SO4 gives 3 ions, so n = 3.
2. Molality is moles per kg of solvent.

Question 9: A solution is made by mixing 8.5 g of \( \ce{CH2Cl2} \) and 11.95 g of \( \ce{CHCl3} \). The vapour pressures at 298 K are 415 mm Hg and 200 mm Hg, respectively. The mole fraction of \( \ce{CHCl3} \) in the vapour is:
(a) 0.675
(b) 0.325
(c) 0.162
(d) 0.486
Show Answer & Explanation

Answer: (b) 0.325

Explanation:
1. Moles: CH2Cl2 8.5/85 = 0.1; CHCl3 11.95/119.5 = 0.1, so each X = 0.5.
2. \( p_1=207.5 \), \( p_2=100 \); \( y_{CHCl_3}=100/307.5=0.325 \).

Teacher's Note:
1. The vapour is richer in the more volatile CH2Cl2 (0.675).
2. Dalton's law of partial pressures.

Question 10: Which statement about Henry's law is NOT correct?
(a) Different gases have different \( K_H \) values at the same temperature
(b) \( K_H \) increases with temperature and depends on the nature of the gas
(c) The partial pressure of the gas is proportional to its mole fraction in the solution
(d) The higher the value of \( K_H \) at a given pressure, the higher is the solubility of the gas
Show Answer & Explanation

Answer: (d) The higher the value of \( K_H \) at a given pressure, the higher is the solubility of the gas

Explanation:
1. \( p=K_HX \), so for a given pressure \( X=p/K_H \).
2. A higher K_H therefore means a lower solubility, so the statement is wrong.

Teacher's Note:
1. Gases like CO2 with small K_H are quite soluble.
2. Warm water holds less gas.

Question 11: Liquids A and B form an ideal solution. At 350 K the vapour pressures of pure A and pure B are \( 7\times10^3 \) Pa and \( 12\times10^3 \) Pa. The composition of the vapour in equilibrium with a solution containing 40 mole per cent A is:
(a) \( y_A=0.28,\ y_B=0.72 \)
(b) \( y_A=0.76,\ y_B=0.24 \)
(c) \( y_A=0.37,\ y_B=0.63 \)
(d) \( y_A=0.4,\ y_B=0.6 \)
Show Answer & Explanation

Answer: (a) \( y_A=0.28,\ y_B=0.72 \)

Explanation:
1. \( p_A=0.4\times7000=2800 \); \( p_B=0.6\times12000=7200 \); total 10000 Pa.
2. \( y_A=0.28 \), \( y_B=0.72 \).

Teacher's Note:
1. B is more volatile, so the vapour is richer in B.
2. y_A + y_B = 1.

Question 12: The elevation in boiling point for a 1 molal glucose solution is 2 K. The depression in freezing point for a 2 molal glucose solution in the same solvent is also 2 K. The relation between \( K_b \) and \( K_f \) is:
(a) \( K_b=0.5K_f \)
(b) \( K_b=K_f \)
(c) \( K_b=1.5K_f \)
(d) \( K_b=2K_f \)
Show Answer & Explanation

Answer: (d) \( K_b=2K_f \)

Explanation:
1. \( K_b\times1=2 \), so K_b = 2 K per molal.
2. \( K_f\times2=2 \), so K_f = 1 K per molal.
3. K_b = 2 K_f.

Teacher's Note:
1. Glucose is a non-electrolyte, i = 1.
2. The same solvent has different Kb and Kf.

Question 13: The freezing point of a diluted milk sample is -0.2 C, while it should be -0.5 C for pure milk. How much water has been added to the pure milk?
(a) 1 cup of water to 3 cups of pure milk
(b) 3 cups of water to 2 cups of pure milk
(c) 1 cup of water to 2 cups of pure milk
(d) 2 cups of water to 3 cups of pure milk
Show Answer & Explanation

Answer: (b) 3 cups of water to 2 cups of pure milk

Explanation:
1. Depression is proportional to concentration: 0.2/0.5 = 0.4, so the diluted milk has 0.4 of the original concentration.
2. The volume therefore became 1/0.4 = 2.5 times the milk volume: 2 cups of milk become 5 cups.
3. Water added = 5 - 2 = 3 cups for every 2 cups of milk.

Teacher's Note:
1. Check: 2 cups milk + 3 cups water = 5 cups; 2/5 = 0.4.
2. Assume the freezing point depression is proportional to the milk content.

Question 14: \( \ce{K2HgI4} \) is 40% ionised in aqueous solution. The value of its van't Hoff factor is:
(a) 1.6
(b) 2.0
(c) 1.8
(d) 2.2
Show Answer & Explanation

Answer: (c) 1.8

Explanation:
1. K2HgI4 gives 2 K+ and [HgI4]2-: n = 3.
2. \( i=1+(3-1)\times0.4=1.8 \).

Teacher's Note:
1. i = 1 + (n - 1) alpha.
2. Complete ionisation would give 3.

Question 15: The freezing point of a 4% aqueous solution of X is equal to that of a 12% aqueous solution of Y. If the molar mass of X is A, the molar mass of Y is:
(a) 4A
(b) 3A
(c) 2A
(d) A
Show Answer & Explanation

Answer: (b) 3A

Explanation:
1. Equal freezing points mean equal molalities (non-electrolytes): \( 4/A=12/M_Y \).
2. \( M_Y=3A \).

Teacher's Note:
1. The percentages are by mass and the solvent masses are almost equal.
2. Use mass percent / molar mass.

Question 16: Molecules of benzoic acid dimerise in benzene. w g of the acid dissolved in 30 g of benzene shows a depression of 2 K. If the percentage association is 80, then w is (\( K_f=5 \) K kg mol\(^{-1}\), M = 122):
(a) 2.4 g
(b) 1.8 g
(c) 1.0 g
(d) 1.5 g
Show Answer & Explanation

Answer: (a) 2.4 g

Explanation:
1. For dimerisation \( i=1-\alpha/2=0.6 \).
2. \( 2=0.6\times5\times m \), so m = 0.667 mol/kg.
3. Moles = 0.667 x 0.030 = 0.020; w = 0.020 x 122 = 2.44 g.

Teacher's Note:
1. Association lowers the number of particles.
2. i = 0.6 here.

Question 17: The vapour pressures of pure liquids A and B are 400 and 600 mm Hg at 298 K. On mixing, the final volume equals the sum of the initial volumes, and the mole fraction of B is 0.5. The vapour pressure of the solution and the mole fractions of A and B in the vapour are:
(a) 450 mm Hg; 0.4 and 0.6
(b) 500 mm Hg; 0.5 and 0.5
(c) 450 mm Hg; 0.5 and 0.5
(d) 500 mm Hg; 0.4 and 0.6
Show Answer & Explanation

Answer: (d) 500 mm Hg; 0.4 and 0.6

Explanation:
1. \( P=0.5\times400+0.5\times600=500 \) mm Hg.
2. \( y_A=200/500=0.4 \), \( y_B=300/500=0.6 \).

Teacher's Note:
1. No volume change shows an ideal solution.
2. The vapour is richer in the more volatile B.

Question 18: The osmotic pressure of a dilute solution of an ionic compound XY is four times that of a 0.01 M \( \ce{BaCl2} \) solution. Assuming complete dissociation, the concentration of XY is:
(a) \( 4\times10^{-2} \) mol/L
(b) \( 16\times10^{-4} \) mol/L
(c) \( 6\times10^{-2} \) mol/L
(d) \( 4\times10^{-4} \) mol/L
Show Answer & Explanation

Answer: (c) \( 6\times10^{-2} \) mol/L

Explanation:
1. BaCl2: i = 3, particle concentration = 0.03 M.
2. XY: i = 2, so \( 2c=4\times0.03=0.12 \), c = 0.06 M.

Teacher's Note:
1. Equal temperatures, so compare i C.
2. XY gives X+ and Y-.

Question 19: The molal depression constant for a solvent is 4.0 K kg mol\(^{-1}\). The depression in freezing point for a 0.03 mol/kg solution of \( \ce{K2SO4} \) (complete dissociation) is:
(a) 0.36 K
(b) 0.12 K
(c) 0.24 K
(d) 0.18 K
Show Answer & Explanation

Answer: (a) 0.36 K

Explanation:
1. i = 3; \( \Delta T_f=3\times4.0\times0.03=0.36 \) K.

Teacher's Note:
1. Without dissociation it would be 0.12 K.
2. K2SO4 gives 3 ions.

Question 20: A dilute solution is made by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water is 35 mm Hg, the lowering of vapour pressure is:
(a) 0.027 mm Hg
(b) 0.028 mm Hg
(c) 0.031 mm Hg
(d) 0.017 mm Hg
Show Answer & Explanation

Answer: (d) 0.017 mm Hg

Explanation:
1. Moles: urea 0.01; water 20.
2. \( \Delta p=35\times\dfrac{0.01}{20.01}=0.0175\approx0.017 \) mm Hg.

Teacher's Note:
1. Relative lowering = mole fraction of solute.
2. Dilute: 0.01/20 = 5 x 10^-4.

Question 21: 1 g of a non-volatile non-electrolyte is dissolved in 100 g of each of two solvents A and B whose ebullioscopic constants are in the ratio 1 : 5. The ratio of the elevations in their boiling points \( (\Delta T_b)_A : (\Delta T_b)_B \) is:
(a) 10 : 1
(b) 1 : 5
(c) 1 : 0.2
(d) 5 : 1
Show Answer & Explanation

Answer: (b) 1 : 5

Explanation:
1. The molality is the same in both solutions.
2. \( \Delta T_b\propto K_b \), so the ratio is 1 : 5.

Teacher's Note:
1. 1 : 0.2 is the same as 5 : 1, the inverse.
2. The solute and solvent masses cancel.

Question 22: A solution is made by dissolving 0.6 g of urea (60 g/mol) and 1.8 g of glucose (180 g/mol) in 100 mL of water at 27 C. The osmotic pressure is (R = 0.08206 L atm K\(^{-1}\) mol\(^{-1}\)):
(a) 8.2 atm
(b) 1.64 atm
(c) 4.92 atm
(d) 2.46 atm
Show Answer & Explanation

Answer: (c) 4.92 atm

Explanation:
1. Moles: urea 0.01, glucose 0.01, total 0.02 in 0.1 L, so C = 0.2 M.
2. \( \pi=0.2\times0.08206\times300=4.92 \) atm.

Teacher's Note:
1. Both are non-electrolytes.
2. Osmotic pressures of the components add.

Question 23: Vapour pressures of mixtures of benzene (\( \ce{C6H6} \)) and toluene at 50 C are given by \( P=179X_B+92 \) mm Hg, where \( X_B \) is the mole fraction of benzene. The vapour pressures of pure toluene and pure benzene are, respectively:
(a) 179 mm Hg and 92 mm Hg
(b) 92 mm Hg and 271 mm Hg
(c) 91 mm Hg and 180 mm Hg
(d) 92 mm Hg and 179 mm Hg
Show Answer & Explanation

Answer: (b) 92 mm Hg and 271 mm Hg

Explanation:
1. At \( X_B=0 \) (pure toluene) \( P=92 \).
2. At \( X_B=1 \) (pure benzene) \( P=179+92=271 \).

Teacher's Note:
1. The line for an ideal solution joins the two pure vapour pressures.
2. The slope equals p_B0 - p_T0 = 179.

Question 24: Using \( P=179X_B+92 \) mm Hg for benzene-toluene at 50 C, the vapour pressure of the liquid mixture obtained by mixing 936 g benzene (M = 78) and 736 g toluene (M = 92) is:
(a) 199.4 mm Hg
(b) 300 mm Hg
(c) 250 mm Hg
(d) 180.6 mm Hg
Show Answer & Explanation

Answer: (a) 199.4 mm Hg

Explanation:
1. Moles: benzene 936/78 = 12; toluene 736/92 = 8.
2. \( X_B=12/20=0.6 \).
3. \( P=179\times0.6+92=199.4 \) mm Hg.

Teacher's Note:
1. Check: 0.6 x 271 + 0.4 x 92 = 199.4.
2. The answer lies between 92 and 271.

Question 25: 1 g of a mixture of glucose and urea in 250 mL of aqueous solution shows an osmotic pressure of 0.74 atm at 27 C. The percentage of urea in the mixture is (R = 0.0821):
(a) 10.0%
(b) 25.0%
(c) 50.0%
(d) 17.6%
Show Answer & Explanation

Answer: (d) 17.6%

Explanation:
1. Total moles = \( \dfrac{\pi V}{RT}=\dfrac{0.74\times0.25}{0.0821\times300}=7.51\times10^{-3} \).
2. Let urea = x g: \( \dfrac{x}{60}+\dfrac{1-x}{180}=7.51\times10^{-3} \), so \( 2x+1=1.352 \) and x = 0.176 g.
3. Percentage of urea = 17.6%.

Teacher's Note:
1. Both solutes are non-electrolytes.
2. Moles add.

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