JEE Chemistry Solutions MCQs Set 21

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Multiple Choice Questions

Question 1: The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be
(a) 18.0
(b) 342
(c) 60
(d) 180
Show Answer & Explanation

Answer: (d) 180

Explanation:
1. Relative lowering = n/(n + N) with N = 1000/18 = 55.56.
2. 0.00713 = n/(n + 55.56) gives n = 0.399 mol.
3. M = 71.5/0.399 = 179 ≈ 180 g/mol.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 2: Benzene and toluene form nearly ideal solutions. At 20 °C the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20 °C for a solution containing 78 g of benzene and 46 g of toluene is
(a) 50
(b) 25
(c) 37.5
(d) 53.5
Show Answer & Explanation

Answer: (a) 50

Explanation:
1. Moles of benzene = 78/78 = 1; moles of toluene = 46/92 = 0.5.
2. x(benzene) = 1/1.5 = 2/3, giving p(benzene) = (2/3) × 75 = 50 torr.

Teacher's Note:
1. Raoult's law: partial pressure = mole fraction × vapour pressure of the pure liquid.
2. Work with moles, not masses.

Question 3: The vapour pressure lowering caused by the addition of 100 g of sucrose (molecular mass 342) to 1000 g of water, if the vapour pressure of pure water at 25 °C is 23.8 mm Hg, is
(a) 1.25 mm Hg
(b) 0.125 mm Hg
(c) 1.15 mm Hg
(d) 0.12 mm Hg
Show Answer & Explanation

Answer: (b) 0.125 mm Hg

Explanation:
1. Moles of sucrose = 100/342 = 0.2924; moles of water = 55.56.
2. x(solute) = 0.2924/55.85 = 0.00524.
3. ΔP = 23.8 × 0.00524 = 0.125 mm Hg.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 4: A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20 °C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be
(a) 0.549
(b) 0.200
(c) 0.786
(d) 0.478
Show Answer & Explanation

Answer: (d) 0.478

Explanation:
1. In the liquid x(pentane) = 0.2 and x(hexane) = 0.8.
2. p(pentane) = 0.2 × 440 = 88; p(hexane) = 0.8 × 120 = 96; total = 184 mm Hg.
3. y(pentane) = 88/184 = 0.478.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 5: An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressures of methanol and ethanol are 2.619 kPa and 4.556 kPa respectively, the composition of the vapour (in mole fraction) will be
(a) 0.635 methanol, 0.365 ethanol
(b) 0.365 methanol, 0.635 ethanol
(c) 0.574 methanol, 0.326 ethanol
(d) 0.173 methanol, 0.827 ethanol
Show Answer & Explanation

Answer: (b) 0.365 methanol, 0.635 ethanol

Explanation:
1. Mole fraction in vapour = partial pressure / total pressure.
2. Total = 2.619 + 4.556 = 7.175 kPa.
3. y(methanol) = 2.619/7.175 = 0.365 and y(ethanol) = 0.635.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 6: When 20 g of a non-ionic substance is dissolved in 100 g of water, the vapour pressure is lowered by 0.30 mm. The vapour pressure of water at 20 °C is 17.54 mm. The molecular weight of the substance is
(a) 210.2
(b) 206.88
(c) 215.2
(d) 200.8
Show Answer & Explanation

Answer: (b) 206.88

Explanation:
1. Relative lowering = 0.30/17.54 = 0.0171 = n/(n + 5.556).
2. n = 0.0967 mol, so M = 20/0.0967 = 206.9 g/mol.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 7: The vapour pressure of pure liquid A is 0.80 atm. On mixing a non-volatile substance B to A, its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is
(a) 0.150
(b) 0.25
(c) 0.50
(d) 0.75
Show Answer & Explanation

Answer: (b) 0.25

Explanation:
1. Relative lowering = (0.80 − 0.60)/0.80 = 0.25.
2. This equals the mole fraction of the solute, so x(B) = 0.25.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 8: At 300 K, when a solute is added to a solvent, its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of the mole fraction of the solute will be
(a) 0.005
(b) 0.010
(c) 0.100
(d) 0.900
Show Answer & Explanation

Answer: (c) 0.100

Explanation:
1. x(solute) = (50 − 45)/50 = 0.10.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 9: The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?
(a) 0.8
(b) 0.6
(c) 0.4
(d) 0.2
Show Answer & Explanation

Answer: (c) 0.4

Explanation:
1. Lowering is proportional to the mole fraction of solute: 10 mm for 0.2, so 20 mm needs x(solute) = 0.4.
2. x(solvent) = 1 − 0.4 = 0.6.

Teacher's Note:
1. Pick the solvent mole fraction, not the solute mole fraction.

Question 10: The Van't Hoff factor i for a dilute solution of Na2SO4 assuming degree of dissociation α is
(a) 1 + α
(b) 1 + 2α
(c) 1 − α
(d) 1 + α/2
Show Answer & Explanation

Answer: (b) 1 + 2α

Explanation:
1. Na2SO4 → 2Na+ + SO42−, so n = 3 ions.
2. i = 1 + (n − 1)α = 1 + 2α.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 11: The osmotic pressure of a solution containing 0.1 mole of solute per litre at 273 K is (in atm)
(a) 0.1 × 0.08205 × 273
(b) 0.1 × 1 × 0.08205 × 273
(c) About 2.24 atm
(d) 0.0224 atm
Show Answer & Explanation

Answer: (c) About 2.24 atm

Explanation:
1. π = CRT = 0.1 × 0.08205 × 273 = 2.24 atm.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 12: Osmotic pressure is 0.0821 atm at a temperature of 300 K. The concentration in mole/litre is
(a) 0.033
(b) 0.066
(c) 0.33 × 10−2
(d) 3
Show Answer & Explanation

Answer: (c) 0.33 × 10−2

Explanation:
1. π = CRT, so C = π/RT = 0.0821/(0.0821 × 300) = 0.00333 mol/L.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 13: A 5% solution of cane sugar (molecular weight = 342) is isotonic with 1% solution of a substance X. The molecular weight of X is
(a) 34.2
(b) 171.2
(c) 68.4
(d) 136.8
Show Answer & Explanation

Answer: (c) 68.4

Explanation:
1. Isotonic solutions have equal molar concentrations: 50/342 = 10/M(X).
2. M(X) = 3420/50 = 68.4.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 14: The average osmotic pressure of human blood is 7.8 bar at 37 °C. What is the concentration of an aqueous NaCl solution that could be used in the blood stream? (R = 0.0831 L bar K−1 mol−1, assume complete ionisation)
(a) 0.16 mol/L
(b) 0.32 mol/L
(c) 0.60 mol/L
(d) 0.45 mol/L
Show Answer & Explanation

Answer: (a) 0.16 mol/L

Explanation:
1. Total particle concentration = π/RT = 7.8/(0.0831 × 310) = 0.303 mol/L.
2. NaCl gives 2 particles, so [NaCl] = 0.303/2 = 0.15 ≈ 0.16 mol/L.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 15: If 20 g of a solute was dissolved in 500 mL of water and the osmotic pressure of the solution was found to be 600 mm of Hg at 15 °C, the molecular weight of the solute is
(a) 1000
(b) 1200
(c) 1400
(d) 1800
Show Answer & Explanation

Answer: (b) 1200

Explanation:
1. π = 600/760 = 0.789 atm.
2. M = wRT/(πV) = 20 × 0.0821 × 288/(0.789 × 0.5) = 1198 ≈ 1200 g/mol.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 16: A solution containing 4.0 g of polyvinyl chloride polymer in 1 litre of dioxane was found to have an osmotic pressure of 6.0 × 10−4 atm at 300 K. The molecular mass of the polymer is (R = 0.082 L atm mol−1 K−1)
(a) 3.0 × 102
(b) 1.6 × 105
(c) 5.6 × 104
(d) 6.4 × 102
Show Answer & Explanation

Answer: (b) 1.6 × 105

Explanation:
1. M = wRT/(πV) = 4.0 × 0.082 × 300/(6.0 × 10−4 × 1) = 1.64 × 105.

Teacher's Note:
1. Osmotic pressure is the best method for the molar mass of polymers because it gives a large measurable value.

Question 17: A solution of urea at 10 °C has an osmotic pressure of 500 mm. The osmotic pressure becomes 105.3 mm when it is diluted and the temperature raised to 25 °C. The extent of dilution is
(a) 6 times
(b) 5 times
(c) 7 times
(d) 4 times
Show Answer & Explanation

Answer: (b) 5 times

Explanation:
1. π = CRT, so C1/C2 = (π1/π2) × (T2/T1) = (500/105.3) × (298/283).
2. = 4.748 × 1.053 = 5.0, so the solution was diluted 5 times.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 18: Urea solution of 8.6 g/litre (molecular weight 60) is isotonic with a 5% solution of a non-volatile solute. The molecular weight of the solute will be
(a) 348.9
(b) 34.89
(c) 3489
(d) 861.2
Show Answer & Explanation

Answer: (a) 348.9

Explanation:
1. Molar concentration of urea = 8.6/60 = 0.1433 mol/L.
2. The 5% solute has 50 g/L, so M = 50/0.1433 = 348.8 g/mol.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 19: If 0.1 M glucose (molecular weight 180) solution and 0.1 M urea (molecular weight 60) solution are placed on the two sides of a semipermeable membrane to equal heights, then it will be correct to say that
(a) There will be no net movement across the membrane
(b) Glucose will flow across the membrane into urea solution
(c) Urea will flow across the membrane into glucose solution
(d) Water will flow from urea solution into glucose solution
Show Answer & Explanation

Answer: (a) There will be no net movement across the membrane

Explanation:
1. Both solutions have the same molarity and so the same osmotic pressure.
2. They are isotonic, so there is no net flow of solvent.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 20: The osmotic pressure of a 1 molal solution at 27 °C (taking the solution as ideal, molality = molarity) is approximately
(a) 2.46 atm
(b) 24.6 atm
(c) 1.21 atm
(d) 12.1 atm
Show Answer & Explanation

Answer: (b) 24.6 atm

Explanation:
1. π = CRT = 1 × 0.0821 × 300 = 24.6 atm.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 21: Two solutions of KNO3 and CH3COOH are prepared separately. The molarity of both is 0.1 M and their osmotic pressures are P1 and P2 respectively. The correct relationship is
(a) P2 > P1
(b) P1 = P2
(c) P1 > P2
(d) P1/(P1 + P2) = P2/(P1 + P2)
Show Answer & Explanation

Answer: (c) P1 > P2

Explanation:
1. KNO3 is a strong electrolyte (i ≈ 2) whereas acetic acid is a weak electrolyte (i just above 1).
2. So P1 > P2.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

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