JEE Chemistry Solutions MCQs Set 07

Multiple Choice Questions (MCQs) for JEE Chemistry: Solutions

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Multiple Choice Questions

Question 1: Henry's law constant for dissolution of CH4 in benzene at 298 K is 2 × 105 mm Hg. The mole fraction solubility of CH4 in benzene at 298 K under 760 mm Hg is
(a) 1.2 × 10−5
(b) 3.8 × 10−3
(c) 4 × 10−7
(d) 1 × 10−2
Show Answer & Explanation

Answer: (b) 3.8 × 10−3

Explanation:
1. p = KH x, so x = 760/(2 × 105) = 3.8 × 10−3.

Teacher's Note:
1. Higher KH means lower solubility.

Question 2: Which of the following gases does not obey Henry's law?
(a) NH3
(b) H2
(c) O2
(d) He
Show Answer & Explanation

Answer: (a) NH3

Explanation:
1. Gases that react with the solvent (NH3 forms NH4OH) do not obey Henry's law.

Teacher's Note:
1. Henry's law applies to non-reacting gases at moderate pressure.

Question 3: 1 mol of heptane (V.P. = 92 mm Hg) was mixed with 4 mol of octane (V.P. = 31 mm Hg). The vapour pressure of the resulting ideal solution is
(a) 46.2 mm Hg
(b) 40.0 mm Hg
(c) 43.2 mm Hg
(d) 38.4 mm Hg
Show Answer & Explanation

Answer: (c) 43.2 mm Hg

Explanation:
1. x(heptane) = 0.2, x(octane) = 0.8.
2. P = 0.2 × 92 + 0.8 × 31 = 18.4 + 24.8 = 43.2 mm Hg.

Teacher's Note:
1. Raoult's law for each component, then add.

Question 4: At 88°C benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. The mole fraction of benzene in the mixture that will boil at 88°C at 1 atm (ideal solution) is
(a) 0.416
(b) 0.588
(c) 0.688
(d) 0.740
Show Answer & Explanation

Answer: (d) 0.740

Explanation:
1. Boiling at 1 atm means total vapour pressure = 760 torr.
2. 900x + 360(1 − x) = 760, so 540x = 400 and x = 0.740.

Teacher's Note:
1. A liquid boils when its vapour pressure equals the external pressure.

Question 5: If PA0 and PB0 are 108 and 36 torr respectively, the mole fraction of A in the vapour phase if B has a mole fraction of 0.5 in solution is
(a) 0.25
(b) 0.75
(c) 0.60
(d) 0.35
Show Answer & Explanation

Answer: (b) 0.75

Explanation:
1. xA = 0.5.
2. pA = 54, pB = 18, total = 72.
3. yA = 54/72 = 0.75.

Teacher's Note:
1. Vapour is richer in the more volatile component.

Question 6: The relation between the mole fraction of A in the vapour phase (YA) and in the solution (XA), with PA0 the vapour pressure of pure A and PS the total vapour pressure, is
(a) \( (1-X_A)P_A^0 \)
(b) \( \dfrac{X_A}{1-X_A}P_A^0 \)
(c) \( \dfrac{1-X_A}{X_A}P_A^0 \)
(d) \( \dfrac{P_A^0X_A}{P_S} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{P_A^0X_A}{P_S} \)

Explanation:
1. Dalton's law: YA = pA/PS.
2. With Raoult's law pA = PA0XA, YA = PA0XA/PS.

Teacher's Note:
1. Combine Raoult's and Dalton's laws.

Question 7: Which condition is not satisfied by an ideal solution?
(a) ΔHmix = 0
(b) ΔVmix = 0
(c) ΔSmix = 0
(d) obeyance of Raoult's law
Show Answer & Explanation

Answer: (c) ΔSmix = 0

Explanation:
1. Mixing always increases randomness, so ΔSmix > 0 even for an ideal solution.
2. ΔH and ΔV are zero.

Teacher's Note:
1. Ideal: ΔH = 0, ΔV = 0, ΔG < 0, ΔS > 0.

Question 8: An azeotropic mixture of two liquids has a boiling point lower than either of them when it
(a) shows a positive deviation from Raoult's law
(b) shows no deviation from Raoult's law
(c) shows a positive deviation from Henry's law
(d) shows a negative deviation from Raoult's law
Show Answer & Explanation

Answer: (a) shows a positive deviation from Raoult's law

Explanation:
1. Positive deviation raises the vapour pressure, giving a minimum-boiling azeotrope.
2. Negative deviation gives a maximum-boiling azeotrope.

Teacher's Note:
1. Positive deviation: weaker A–B attraction, e.g. ethanol-water.

Question 9: The relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute. This statement was given by
(a) Raoult
(b) Henry
(c) Joule
(d) Dalton
Show Answer & Explanation

Answer: (a) Raoult

Explanation:
1. This is Raoult's law for dilute solutions of non-volatile solutes.

Teacher's Note:
1. Henry's law concerns gas solubility.

Question 10: Which one of the following is the incorrect form of Raoult's law? (P0 = pure solvent, Ps = solution, n = moles of solute, N = moles of solvent)
(a) \( \dfrac{P_s}{P^0} = \dfrac{N}{n+N} \)
(b) \( \dfrac{P^0}{P^0-P_s} = 1 + \dfrac{N}{n} \)
(c) \( \dfrac{P^0-P_s}{P_s} = \dfrac{n}{n+N} \)
(d) \( \dfrac{P_s}{P^0-P_s} = \dfrac{N}{n} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{P^0-P_s}{P_s} = \dfrac{n}{n+N} \)

Explanation:
1. Correct form: (P0 − Ps)/Ps = n/N.
2. The version with n/(n + N) in the denominator belongs to (P0 − Ps)/P0, so (c) is wrong.

Teacher's Note:
1. The three correct forms follow from (P0 − Ps)/P0 = n/(n + N).

Question 11: The vapour pressure of a solution having a solid as solute and a liquid as solvent is
(a) directly proportional to the mole fraction of the solvent
(b) inversely proportional to the mole fraction of the solvent
(c) directly proportional to the mole fraction of the solute
(d) inversely proportional to the mole fraction of the solute
Show Answer & Explanation

Answer: (a) directly proportional to the mole fraction of the solvent

Explanation:
1. Ps = xsolvent P0.

Teacher's Note:
1. Raoult's law for the solvent.

Question 12: If P0 and PS are the vapour pressures of the solvent and its solution, and N1 and N2 are the mole fractions of the solvent and solute respectively, then
(a) \( P_S = \dfrac{P_0}{N_2} \)
(b) \( P_0 - P_S = P_0N_2 \)
(c) \( P_S = P_0N_2 \)
(d) \( \dfrac{P_0-P_S}{P_S} = \dfrac{N_1}{N_1+N_2} \)
Show Answer & Explanation

Answer: (b) \( P_0 - P_S = P_0N_2 \)

Explanation:
1. PS = P0N1 = P0(1 − N2), so P0 − PS = P0N2.

Teacher's Note:
1. The lowering of vapour pressure is proportional to the mole fraction of solute.

Question 13: One mole of a non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is
(a) \( \dfrac23 \)
(b) \( \dfrac13 \)
(c) \( \dfrac12 \)
(d) \( \dfrac32 \)
Show Answer & Explanation

Answer: (a) \( \dfrac23 \)

Explanation:
1. x(water) = 2/3.
2. Ps/P0 = 2/3.

Teacher's Note:
1. Relative vapour pressure equals the mole fraction of solvent.

Question 14: The vapour pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373 K. The mole fraction of solute is
(a) \( \dfrac1{10} \)
(b) \( \dfrac1{7.6} \)
(c) \( \dfrac1{35} \)
(d) \( \dfrac1{76} \)
Show Answer & Explanation

Answer: (d) \( \dfrac1{76} \)

Explanation:
1. At 373 K pure water has P0 = 760 mm.
2. xsolute = (760 − 750)/760 = 1/76.

Teacher's Note:
1. Water boils at 373 K, so its vapour pressure is 760 mm Hg.

Question 15: The vapour pressure of water at room temperature is 23.8 mm Hg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is equal to
(a) 23.9 mm Hg
(b) 24.2 mm Hg
(c) 21.42 mm Hg
(d) 31.44 mm Hg
Show Answer & Explanation

Answer: (c) 21.42 mm Hg

Explanation:
1. x(water) = 0.9.
2. P = 0.9 × 23.8 = 21.42 mm Hg.

Teacher's Note:
1. Do not use the solute mole fraction directly.

Question 16: The vapour pressure of pure A is 10 torr and at the same temperature, when 1 g of B is dissolved in 20 g of A, its vapour pressure is reduced to 9.0 torr. If the molecular mass of A is 200 amu, then the molecular mass of B is
(a) 100 amu
(b) 90 amu
(c) 75 amu
(d) 120 amu
Show Answer & Explanation

Answer: (b) 90 amu

Explanation:
1. (P0 − Ps)/Ps = nB/nA = 1/9.
2. nA = 20/200 = 0.1, so nB = 0.1/9 = 0.0111 mol.
3. MB = 1/0.0111 = 90.

Teacher's Note:
1. Use the exact form (P0 − Ps)/Ps = n/N.

Question 17: The boiling points of C6H6, CH3OH, C6H5NH2 and C6H5NO2 are 80°C, 65°C, 184°C and 212°C respectively. Which will show the highest vapour pressure at room temperature?
(a) C6H6
(b) CH3OH
(c) C6H5NH2
(d) C6H5NO2
Show Answer & Explanation

Answer: (b) CH3OH

Explanation:
1. The lowest boiling liquid is the most volatile and has the highest vapour pressure at a given temperature.
2. CH3OH boils at 65°C.

Teacher's Note:
1. Low boiling point means high vapour pressure at room temperature.

Question 18: If Raoult's law is obeyed, the vapour pressure of the solvent in a solution is directly proportional to
(a) mole fraction of the solvent
(b) mole fraction of the solute
(c) mole fraction of the solvent and solute
(d) the volume of the solution
Show Answer & Explanation

Answer: (a) mole fraction of the solvent

Explanation:
1. p = xsolvent P0.

Teacher's Note:
1. The proportionality constant is the vapour pressure of the pure solvent.

Question 19: The vapour pressure of a pure liquid A is 70 torr at 27°C. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and the total vapour pressure of the solution is 84 torr at 27°C. The vapour pressure of pure liquid B at 27°C is
(a) 14
(b) 56
(c) 140
(d) 70
Show Answer & Explanation

Answer: (c) 140

Explanation:
1. 84 = 0.8 × 70 + 0.2 PB0 = 56 + 0.2 PB0.
2. PB0 = 28/0.2 = 140 torr.

Teacher's Note:
1. Total pressure = sum of partial pressures.

Question 20: PA and PB are the vapour pressures of pure liquid components A and B of an ideal binary solution. If xA is the mole fraction of A, the total vapour pressure of the solution will be
(a) PA + xA(PB − PA)
(b) PB + xA(PA − PB)
(c) PA + xB(PA − PB)
(d) PB + xB(PB − PA)
Show Answer & Explanation

Answer: (b) PB + xA(PA − PB)

Explanation:
1. P = xAPA + (1 − xA)PB = PB + xA(PA − PB).

Teacher's Note:
1. Eliminate xB by xB = 1 − xA.

Question 21: The vapour pressures of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25°C are 200 mm Hg and 415 mm Hg. The vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be (molar masses 119.5 and 85)
(a) 347.9 mm Hg
(b) 280.5 mm Hg
(c) 173.9 mm Hg
(d) 615 mm Hg
Show Answer & Explanation

Answer: (a) 347.9 mm Hg

Explanation:
1. n(CHCl3) = 25.5/119.5 = 0.2134; n(CH2Cl2) = 40/85 = 0.4706.
2. x = 0.3119 and 0.6881.
3. P = 0.3119 × 200 + 0.6881 × 415 = 62.4 + 285.5 = 347.9 mm Hg.

Teacher's Note:
1. Total pressure lies between the two pure values.

Question 22: For an ideal solution of A and B, which statement is incorrect?
(a) The enthalpy change of mixing is zero
(b) The volume change of mixing is zero
(c) The intermolecular forces of A and B are the same as those of A–A and B–B
(d) The entropy change of mixing is zero
Show Answer & Explanation

Answer: (d) The entropy change of mixing is zero

Explanation:
1. Entropy increases on mixing even in an ideal solution.
2. The other three statements describe an ideal solution.

Teacher's Note:
1. Ideal solution: A–B interactions equal A–A and B–B interactions.

Question 23: Which one is not equal to zero for an ideal solution?
(a) ΔSmix
(b) ΔVmix
(c) ΔP = Pobserved − PRaoult
(d) ΔHmix
Show Answer & Explanation

Answer: (a) ΔSmix

Explanation:
1. ΔV, ΔH and the deviation ΔP are all zero for an ideal solution.
2. ΔSmix is positive.

Teacher's Note:
1. The only non-zero property is entropy.

Question 24: Which of the following is incorrect for an ideal solution?
(a) ΔP = Pobs − Pcalculated by Raoult's law = 0
(b) ΔGmix = 0
(c) ΔHmix = 0
(d) ΔUmix = 0
Show Answer & Explanation

Answer: (b) ΔGmix = 0

Explanation:
1. For spontaneous mixing ΔGmix = −TΔSmix is negative, not zero.

Teacher's Note:
1. ΔGmix < 0 for all spontaneous mixing.

Question 25: A solution of acetone in ethanol
(a) shows a positive deviation from Raoult's law
(b) behaves like a near ideal solution
(c) obeys Raoult's law
(d) shows a negative deviation from Raoult's law
Show Answer & Explanation

Answer: (a) shows a positive deviation from Raoult's law

Explanation:
1. Acetone breaks the hydrogen bonds of ethanol, weakening interactions and raising the vapour pressure above the Raoult value.

Teacher's Note:
1. Positive deviation corresponds to ΔHmix > 0 and ΔVmix > 0.

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Multiple Choice Questions (MCQs) for JEE Chemistry Solutions

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