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Molar-mass determination, abnormal molar mass and van’t Hoff factor: NEET MCQs with answers

70 questions on this topic of Solutions (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following colligative properties is most commonly used for the determination of the molar mass of polymers and biomolecules like proteins?

  1. Depression in freezing point
  2. Elevation in boiling point
  3. Osmotic pressure
  4. Relative lowering of vapour pressure
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Answer: (C) Osmotic pressure

Biomolecules and polymers have poor solubility and undergo decomposition at elevated temperatures. Osmotic pressure measurement is carried out at room temperature and produces measurable values even for extremely dilute solutions of high molar mass solutes.

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2. Assuming complete dissociation of any electrolyte, the depression in freezing point of a 0.25 molal aqueous solution of K₂SO₄ is (K_f of water = 1.86 K kg mol⁻¹):

  1. 1.86 K
  2. 7.44 K
  3. 0.465 K
  4. 1.4 K
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Answer: (D) 1.4 K

ΔT_f = i K_f m = 3 × 1.86 × 0.25 = 1.4 K (van't Hoff factor i = 3).

hard

3. A compound with normal molar mass of 244 g mol⁻¹ dissolves in an organic solvent and undergoes association: 2B ⇌ B₂. When 2.44 g of this solute is dissolved in 50 g of the solvent (K_f = 5.0 K kg mol⁻¹), the observed depression in freezing point is 0.60 K. What is the percentage association of the compound in this solvent?

  1. 60%
  2. 80%
  3. 75%
  4. 40%
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Answer: (B) 80%

Molality m = (2.44 / 244) / (50 / 1000) = 0.01 / 0.05 = 0.20 mol kg⁻¹. ΔT_f = i × K_f × m ⇒ 0.60 = i × 5.0 × 0.20 ⇒ 0.60 = 1.0 × i ⇒ i = 0.60. For dimerization, i = 1 - α/2. Thus, 0.60 = 1 - α/2 ⇒ α/2 = 0.40 ⇒ α = 0.80, which corresponds to 80% association.

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4. The freezing point of an aqueous solution containing 0.05 mol kg⁻¹ of a weak acid HX is -0.106 °C. Calculate the van’t Hoff factor (i) and the degree of dissociation (α) of the acid. (K_f for water = 1.86 K kg mol⁻¹)

  1. i = 1.28, α = 0.28
  2. i = 1.14, α = 0.14
  3. i = 1.07, α = 0.07
  4. i = 1.20, α = 0.20
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Answer: (B) i = 1.14, α = 0.14

ΔT_f = i × K_f × m ⇒ 0.106 = i × 1.86 × 0.05 ⇒ 0.106 = i × 0.093 ⇒ i = 0.106 / 0.093 ≈ 1.14. For a monobasic acid HX ⇌ H⁺ + X⁻ (n = 2), α = i - 1 = 1.14 - 1 = 0.14.

easy

5. Which of the following pairs of 0.10 mol kg⁻¹ aqueous solutions will have the identical elevation in boiling point, assuming complete dissociation of all ionic solutes?

  1. NaCl and Na₂SO₄
  2. AlCl₃ and K₄[Fe(CN)₆]
  3. FeCl₃ and BaCl₂
  4. CaCl₂ and K₂SO₄
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Answer: (D) CaCl₂ and K₂SO₄

ΔT_b is proportional to i × m. For CaCl₂, i = 3 (Ca²⁺ + 2Cl⁻), so i × m = 0.30. For K₂SO₄, i = 3 (2K⁺ + SO₄²⁻), so i × m = 0.30. Since both produce 3 ions upon complete dissociation at equal molality, their boiling point elevations will be identical.

hard

6. An aqueous solution contains a complex represented by Co(NH₃)₄Cl₃. A 0.01 m aqueous solution of this complex freezes at -0.0372 °C. Assuming 100% ionization of the complex and K_f for water = 1.86 K kg mol⁻¹, what is the correct coordination formula of the complex?

  1. [Co(NH₃)₄]Cl₃
  2. [Co(NH₃)₄Cl]Cl₂
  3. [Co(NH₃)₄Cl₃]
  4. [Co(NH₃)₄Cl₂]Cl
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Answer: (D) [Co(NH₃)₄Cl₂]Cl

ΔT_f = i × K_f × m ⇒ 0.0372 = i × 1.86 × 0.01 ⇒ 0.0372 = i × 0.0186 ⇒ i = 2. Since the complex is 100% ionized and i = 2, it must dissociate into 2 ions per formula unit. The formulation [Co(NH₃)₄Cl₂]Cl ionizes into [Co(NH₃)₄Cl₂]⁺ and Cl⁻ (2 ions).

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7. Assuming complete dissociation of any electrolyte, the depression in freezing point of a 0.2 molal aqueous solution of NaCl is (K_f of water = 1.86 K kg mol⁻¹):

  1. 0.744 K
  2. 1.12 K
  3. 0.372 K
  4. 9.3 K
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Answer: (A) 0.744 K

ΔT_f = i K_f m = 2 × 1.86 × 0.2 = 0.744 K (van't Hoff factor i = 2).

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8. Given below are two statements: Statement I: The abnormal molar mass determined from colligative properties is always greater than the normal molar mass when the solute dissociates in the solution. Statement II: Dissociation of solute particles leads to an increase in the total number of particles, which causes an increase in the observed value of colligative properties. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (C) Statement I is incorrect but Statement II is correct

When a solute dissociates, the number of particles increases, leading to a higher colligative property value and consequently a smaller (lower) abnormal molar mass than the theoretical value (M_obs = M_normal / i). Thus, Statement I is incorrect and Statement II is correct.

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9. Assuming complete dissociation of any electrolyte, the depression in freezing point of a 0.25 molal aqueous solution of CaCl₂ is (K_f of water = 1.86 K kg mol⁻¹):

  1. 7.44 K
  2. 1.86 K
  3. 0.465 K
  4. 1.4 K
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Answer: (D) 1.4 K

ΔT_f = i K_f m = 3 × 1.86 × 0.25 = 1.4 K (van't Hoff factor i = 3).

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10. Assuming complete dissociation of any electrolyte, the depression in freezing point of a 0.5 molal aqueous solution of K₂SO₄ is (K_f of water = 1.86 K kg mol⁻¹):

  1. 5.58 K
  2. 3.72 K
  3. 2.79 K
  4. 0.93 K
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Answer: (C) 2.79 K

ΔT_f = i K_f m = 3 × 1.86 × 0.5 = 2.79 K (van't Hoff factor i = 3).

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11. A 0.1 m solution of a weak monobasic acid HA has a freezing point of -0.2046 °C. If the cryoscopic constant (K_f) of water is 1.86 K kg mol⁻¹, the degree of dissociation (α) of the acid is:

  1. 0.10
  2. 0.05
  3. 0.15
  4. 0.20
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Answer: (A) 0.10

ΔT_f = i × K_f × m, so 0.2046 = i × 1.86 × 0.1, giving i = 1.10. For a monobasic acid HA ⇌ H⁺ + A⁻ (n = 2), i = 1 + α. Therefore, α = i - 1 = 1.10 - 1 = 0.10.

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12. Assuming complete dissociation of any electrolyte, the depression in freezing point of a 1 molal aqueous solution of urea is (K_f of water = 1.86 K kg mol⁻¹):

  1. 1.86 K
  2. 0.93 K
  3. 3.72 K
  4. 2.79 K
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Answer: (A) 1.86 K

ΔT_f = i K_f m = 1 × 1.86 × 1 = 1.86 K (van't Hoff factor i = 1).

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