JEE Chemistry Solutions MCQs Set 33

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Multiple Choice Questions

Question 1: A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of the unknown solute in g/mol is:
(a) 136.2
(b) 171.2
(c) 68.4
(d) 34.2
Show Answer & Explanation

Answer: (c) 68.4

Explanation:
1. Isotonic solutions have equal molar concentration: (50 g/L)/342 = (10 g/L)/M.
2. M = 342 × 10/50 = 68.4 g/mol.

Teacher's Note:
1. Treat the percentages as mass per volume.
2. Equal concentration implies equal osmotic pressure.

Question 2: If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (ΔT_f), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (K_f = 1.86 K kg mol⁻¹):
(a) 0.0372 K
(b) 0.0558 K
(c) 0.0744 K
(d) 0.0186 K
Show Answer & Explanation

Answer: (b) 0.0558 K

Explanation:
1. Na₂SO₄ → 2Na⁺ + SO₄²⁻, so i = 3.
2. ΔT_f = 3 × 1.86 × 0.01 = 0.0558 K.

Teacher's Note:
1. Include the van't Hoff factor for the electrolyte.
2. Molality = 0.01 mol kg⁻¹.

Question 3: On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa, respectively. The vapour pressure of the solution obtained by mixing 25 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol⁻¹ and of octane = 114 g mol⁻¹):
(a) 72.0 kPa
(b) 36.1 kPa
(c) 96.2 kPa
(d) 144.5 kPa
Show Answer & Explanation

Answer: (a) 72.0 kPa

Explanation:
1. Moles of heptane = 0.25; octane = 35/114 = 0.307.
2. x_heptane = 0.25/0.557 = 0.45; x_octane = 0.55.
3. p = 0.45 × 105 + 0.55 × 45 = 47.3 + 24.8 = 72.0 kPa.

Teacher's Note:
1. Apply Raoult's law to both components and add.
2. Convert masses to mole fractions.

Question 4: A binary liquid solution is made by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution?
(a) The solution formed is an ideal solution
(b) The solution is non-ideal, showing positive deviation from Raoult's law
(c) The solution is non-ideal, showing negative deviation from Raoult's law
(d) n-heptane shows positive deviation while ethanol shows negative deviation from Raoult's law
Show Answer & Explanation

Answer: (b) The solution is non-ideal, showing positive deviation from Raoult's law

Explanation:
1. Ethanol molecules are strongly hydrogen-bonded; n-heptane disrupts this.
2. The heptane-ethanol attraction is weaker than the ethanol-ethanol attraction, so the escaping tendency rises and both show positive deviation.

Teacher's Note:
1. Weaker A–B interactions give positive deviation.
2. Positive deviation means a higher vapour pressure than predicted.

Question 5: At 80°C, the vapour pressure of pure liquid A is 520 mmHg and that of pure liquid B is 1000 mmHg. If a mixture solution of A and B boils at 80°C and 1 atm pressure, the amount of A in the mixture is (1 atm = 760 mmHg):
(a) 52 mole percent
(b) 34 mole percent
(c) 48 mole percent
(d) 50 mole percent
Show Answer & Explanation

Answer: (d) 50 mole percent

Explanation:
1. Boiling occurs when the total vapour pressure equals 760 mmHg.
2. 760 = 520x_A + 1000(1 − x_A), so 480x_A = 240 and x_A = 0.5.
3. The mixture contains 50 mole percent A.

Teacher's Note:
1. Boiling at 1 atm means p_total = 760 mmHg.
2. Use x_B = 1 − x_A.

Question 6: The density (in g mL⁻¹) of a 3.60 M sulphuric acid solution that is 29% H₂SO₄ (molar mass = 98 g mol⁻¹) by mass will be:
(a) 1.64
(b) 1.88
(c) 1.22
(d) 1.45
Show Answer & Explanation

Answer: (c) 1.22

Explanation:
1. Molarity = 10 × density × (% by mass)/molar mass.
2. 3.60 = 10 × d × 29/98, so d = 3.60 × 98/290 = 1.216 ≈ 1.22 g mL⁻¹.

Teacher's Note:
1. The formula M = 10dx/M₀ links molarity and percentage by mass.
2. Alternatively, 1 L contains 352.8 g H₂SO₄, which is 29% of the mass.

Question 7: A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be:
(a) 350
(b) 300
(c) 700
(d) 360
Show Answer & Explanation

Answer: (a) 350

Explanation:
1. Propyl alcohol has x = 0.4: 290 = 200 × 0.4 + p°_ethyl × 0.6.
2. p°_ethyl = (290 − 80)/0.6 = 350 mm.

Teacher's Note:
1. Apply Raoult's law to the ideal mixture.
2. The question asks for the vapour pressure of pure ethyl alcohol.

Question 8: A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol⁻¹) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm⁻³, molar mass of the substance will be:
(a) 90.0 g mol⁻¹
(b) 115.0 g mol⁻¹
(c) 105.0 g mol⁻¹
(d) 210.0 g mol⁻¹
Show Answer & Explanation

Answer: (d) 210.0 g mol⁻¹

Explanation:
1. Isotonic solutions have the same molar concentration: (5.25/M) = (1.5/60) for equal volumes.
2. M = 5.25 × 60/1.5 = 210 g mol⁻¹.

Teacher's Note:
1. Equal densities allow percentages to be compared directly.
2. Equal osmotic pressure means equal molar concentration.

Question 9: The density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is:
(a) 1.14 mol kg⁻¹
(b) 3.28 mol kg⁻¹
(c) 2.28 mol kg⁻¹
(d) 0.44 mol kg⁻¹
Show Answer & Explanation

Answer: (c) 2.28 mol kg⁻¹

Explanation:
1. In 1 L: mass of solution = 1020 g; acetic acid = 2.05 × 60 = 123 g, so water = 897 g.
2. Molality = 2.05/0.897 = 2.28 mol kg⁻¹.

Teacher's Note:
1. Find the mass of solvent from the density and the mass of solute.
2. Molality is moles per kilogram of solvent.

Question 10: Equimolar solutions of non-electrolytes in the same solvent have:
(a) different boiling point and different freezing point
(b) same boiling point and same freezing point
(c) different boiling point but same freezing point
(d) same boiling point but different freezing point
Show Answer & Explanation

Answer: (b) same boiling point and same freezing point

Explanation:
1. ΔT_b = K_b m and ΔT_f = K_f m depend only on the solvent constants and molality for non-electrolytes.
2. Equimolar solutions have the same boiling point and the same freezing point.

Teacher's Note:
1. Colligative properties depend on the number of solute particles, not their nature.
2. Electrolytes would differ because of the van't Hoff factor.

Question 11: If α is the degree of dissociation of Na₂SO₄, the van't Hoff factor (i) used for calculating the molecular mass is:
(a) 1 − 2α
(b) 1 + 2α
(c) 1 − α
(d) 1 + α
Show Answer & Explanation

Answer: (b) 1 + 2α

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

Teacher's Note:
1. i = 1 + (n − 1)α.
2. For complete dissociation (α = 1) i = 3.

Question 12: 6.02 × 10²⁰ molecules of urea are present in 100 mL of its solution. The concentration of the urea solution is (Avogadro constant N_A = 6.02 × 10²³ mol⁻¹):
(a) 0.001 M
(b) 0.01 M
(c) 0.02 M
(d) 0.1 M
Show Answer & Explanation

Answer: (b) 0.01 M

Explanation:
1. Moles = 6.02 × 10²⁰/6.02 × 10²³ = 0.001 mol.
2. Concentration = 0.001 mol/0.1 L = 0.01 M.

Teacher's Note:
1. Convert molecules to moles using N_A.
2. Volume must be in litres.

Question 13: Which of the following liquid pairs shows a positive deviation from Raoult's law?
(a) Water-hydrochloric acid
(b) Benzene-methanol
(c) Water-nitric acid
(d) Acetone-chloroform
Show Answer & Explanation

Answer: (b) Benzene-methanol

Explanation:
1. Water-HCl and water-HNO₃ show negative deviation (strong interaction), as do acetone-chloroform (hydrogen bonding).
2. Methanol is associated by hydrogen bonds, which benzene breaks, so benzene-methanol shows positive deviation.

Teacher's Note:
1. Weaker A–B interactions than A–A and B–B give positive deviation.
2. Maximum-boiling azeotropes show negative deviation.

Question 14: Which one of the following aqueous solutions will exhibit the highest boiling point?
(a) 0.01 M Na₂SO₄
(b) 0.01 M KNO₃
(c) 0.015 M urea
(d) 0.015 M glucose
Show Answer & Explanation

Answer: (a) 0.01 M Na₂SO₄

Explanation:
1. ΔT_b ∝ i × m: Na₂SO₄: 3 × 0.01 = 0.03; KNO₃: 2 × 0.01 = 0.02; urea and glucose: 0.015.
2. Na₂SO₄ has the highest total particle concentration, so the highest boiling point.

Teacher's Note:
1. Compare the total molar concentration of particles.
2. Non-electrolytes have i = 1.

Question 15: 25 mL of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was:
(a) 0.07
(b) 0.14
(c) 0.28
(d) 0.35
Show Answer & Explanation

Answer: (a) 0.07

Explanation:
1. Ba(OH)₂ is diacidic: normality = 2M. Equivalents: 2M × 25 = 0.1 × 35.
2. M = 3.5/50 = 0.07.

Teacher's Note:
1. Use N₁V₁ = N₂V₂ with N = n × M.
2. Barium hydroxide gives two OH⁻ per formula unit.

Question 16: If liquids A and B form an ideal solution, the:
(a) enthalpy of mixing is zero
(b) entropy of mixing is zero
(c) free energy of mixing is zero
(d) free energy as well as the entropy of mixing are each zero
Show Answer & Explanation

Answer: (a) enthalpy of mixing is zero

Explanation:
1. For an ideal solution ΔH_mix = 0 and ΔV_mix = 0 because A–B interactions equal A–A and B–B interactions.
2. ΔS_mix > 0 and ΔG_mix = −TΔS_mix < 0, so only the enthalpy of mixing is zero.

Teacher's Note:
1. Mixing always increases the entropy.
2. Ideal behaviour means no heat is evolved or absorbed.

Question 17: In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionisation is 0.3. Taking K_f for water as 1.85, the freezing point of the solution will be nearest to:
(a) −0.480°C
(b) −0.360°C
(c) −0.260°C
(d) +0.480°C
Show Answer & Explanation

Answer: (a) −0.480°C

Explanation:
1. HX ⇌ H⁺ + X⁻, so i = 1 + α = 1.3.
2. ΔT_f = 1.3 × 1.85 × 0.2 = 0.481, so the freezing point is −0.481°C, nearest to −0.480°C.

Teacher's Note:
1. Weak acids partly ionise, so i lies between 1 and 2.
2. The freezing point is below 0°C.

Question 18: Which of the following concentration factors is affected by change in temperature?
(a) Molarity
(b) Molality
(c) Mole fraction
(d) Weight fraction
Show Answer & Explanation

Answer: (a) Molarity

Explanation:
1. Molarity is moles per unit volume and the volume of the solution changes with temperature.
2. Molality, mole fraction and weight fraction involve only masses or moles.

Teacher's Note:
1. Mass-based units do not change with temperature.
2. Volume-based units do.

Question 19: For an aqueous solution, the freezing point is −0.186°C. The elevation of the boiling point of the same solution is (K_f = 1.86 K kg mol⁻¹ and K_b = 0.512 K kg mol⁻¹):
(a) 0.186°
(b) 0.0512°
(c) 1.86°
(d) 5.12°
Show Answer & Explanation

Answer: (b) 0.0512°

Explanation:
1. ΔT_f = K_f m gives m = 0.186/1.86 = 0.1 mol kg⁻¹.
2. ΔT_b = K_b m = 0.512 × 0.1 = 0.0512°.

Teacher's Note:
1. The same molality is used for both properties.
2. Find molality from the freezing point depression.

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