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Colligative properties: NEET MCQs with answers

66 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 0.05 m aqueous solutions will have the highest boiling point?

  1. 0.05 m Al₂(SO₄)₃
  2. 0.05 m Na₃PO₄
  3. 0.05 m BaCl₂
  4. 0.05 m Urea
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Answer: (A) 0.05 m Al₂(SO₄)₃

Elevation in boiling point is given by ΔTb = i · Kb · m. For the same molality and solvent, ΔTb is directly proportional to the van 't Hoff factor (i). For complete dissociation: Al₂(SO₄)₃ gives i = 5, Na₃PO₄ gives i = 4, BaCl₂ gives i = 3, and urea gives i = 1. Therefore, 0.05 m Al₂(SO₄)₃ has the highest value of i·m and consequently the highest boiling point.

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2. The osmotic pressure of a 0.5 M aqueous solution of a non-electrolyte at 300 K is (R = 0.0821 L atm K⁻¹ mol⁻¹):

  1. 12.3 atm
  2. 49.3 atm
  3. 1250 atm
  4. 0.0411 atm
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Answer: (A) 12.3 atm

For a non-electrolyte (i = 1), osmotic pressure π = CRT = 0.5 × 0.0821 × 300 = 12.3 atm.

hard

3. A complex of cobalt has the molecular formula CoCl₃·5NH₃. A 0.01 m aqueous solution of this complex exhibits a freezing point depression of 0.0558 K. If Kf for water is 1.86 K kg mol⁻¹ and the complex is 100% ionized, what is the correct coordination formula of the complex?

  1. [Co(NH₃)₅Cl]Cl₂
  2. [Co(NH₃)₄Cl₂]Cl·NH₃
  3. [Co(NH₃)₅]Cl₃
  4. [Co(NH₃)₃Cl₃]·2NH₃
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Answer: (A) [Co(NH₃)₅Cl]Cl₂

ΔTf = i × Kf × m ⇒ 0.0558 = i × 1.86 × 0.01 ⇒ i = 0.0558 / 0.0186 = 3. Since the complex is 100% ionized, i = 3 indicates that each formula unit dissociates into 3 ions (one complex cation and two chloride anions). Therefore, the formula is [Co(NH₃)₅Cl]Cl₂, which gives [Co(NH₃)₅Cl]²⁺ and 2 Cl⁻.

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4. The osmotic pressure of a 0.1 M aqueous solution of a non-electrolyte at 300 K is (R = 0.0821 L atm K⁻¹ mol⁻¹):

  1. 249 atm
  2. 0.00821 atm
  3. 246 atm
  4. 2.46 atm
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Answer: (D) 2.46 atm

For a non-electrolyte (i = 1), osmotic pressure π = CRT = 0.1 × 0.0821 × 300 = 2.46 atm.

easy

5. What is the value of the van 't Hoff factor (i) for a weak electrolyte AB₂ that undergoes 40% dissociation in an aqueous solution?

  1. 1.40
  2. 2.60
  3. 2.20
  4. 1.80
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Answer: (D) 1.80

For the dissociation AB₂ ⇌ A²⁺ + 2B⁻, the number of particles per formula unit is n = 3. The van 't Hoff factor is calculated using i = 1 + (n - 1)α. Given α = 0.40, i = 1 + (3 - 1)(0.40) = 1 + 2(0.40) = 1.80.

hard

6. A 0.02 m aqueous solution of a monobasic acid (HA) freezes at -0.04185 °C. If the cryoscopic constant (Kf) for water is 1.86 K kg mol⁻¹, what is the degree of dissociation (α) of the acid?

  1. 0.500
  2. 0.250
  3. 0.050
  4. 0.125
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Answer: (D) 0.125

Observed depression ΔTf = 0.04185 K. Theoretical ΔTf = Kf · m = 1.86 × 0.02 = 0.0372 K. Therefore, van 't Hoff factor i = ΔTf(obs) / ΔTf(calc) = 0.04185 / 0.0372 = 1.125. For HA ⇌ H⁺ + A⁻, i = 1 + α. Hence, α = 1.125 - 1 = 0.125.

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7. The osmotic pressure of a 0.05 M aqueous solution of a non-electrolyte at 273 K is (R = 0.0821 L atm K⁻¹ mol⁻¹):

  1. 448 atm
  2. 1.12 atm
  3. 113 atm
  4. 0.00411 atm
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Answer: (B) 1.12 atm

For a non-electrolyte (i = 1), osmotic pressure π = CRT = 0.05 × 0.0821 × 273 = 1.12 atm.

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8. The osmotic pressure of a 1 M aqueous solution of a non-electrolyte at 273 K is (R = 0.0821 L atm K⁻¹ mol⁻¹):

  1. 0.0821 atm
  2. 44.8 atm
  3. 22.4 atm
  4. 2270 atm
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Answer: (C) 22.4 atm

For a non-electrolyte (i = 1), osmotic pressure π = CRT = 1 × 0.0821 × 273 = 22.4 atm.

hard

9. A 0.01 m aqueous solution of potassium ferricyanide, K₃[Fe(CN)₆], freezes at -0.06882 °C. What is the apparent degree of dissociation of the salt? (Kf for water = 1.86 K kg mol⁻¹)

  1. 0.80
  2. 0.90
  3. 0.75
  4. 0.85
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Answer: (B) 0.90

ΔTf = i × Kf × m. 0.06882 = i × 1.86 × 0.01, so i = 0.06882 / 0.0186 = 3.70. For K₃[Fe(CN)₆], n = 4. The degree of dissociation α = (i - 1) / (n - 1) = (3.70 - 1) / (4 - 1) = 2.70 / 3 = 0.90.

hard

10. A 1.00 m aqueous solution of glucose boils at 100.52 °C under 1 atm pressure. If 1.00 mole of sodium chloride (NaCl) is dissolved in 1.00 kg of water instead, it boils at 100.98 °C under the same pressure. What is the apparent degree of dissociation of NaCl in this solution? (Kb for water = 0.52 K kg mol⁻¹)

  1. 0.78
  2. 0.82
  3. 0.92
  4. 0.88
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Answer: (D) 0.88

For the NaCl solution, ΔTb = 100.98 - 100.00 = 0.98 °C = 0.98 K. Using ΔTb = i · Kb · m: 0.98 = i × 0.52 × 1.00, which gives i = 0.98 / 0.52 ≈ 1.8846. For NaCl ⇌ Na⁺ + Cl⁻ (n = 2), the degree of dissociation is α = (i - 1) / (n - 1) = (1.8846 - 1) / (2 - 1) ≈ 0.8846 ≈ 0.88 (or 88%).

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11. The osmotic pressure of a 0.1 M aqueous solution of a non-electrolyte at 273 K is (R = 0.0821 L atm K⁻¹ mol⁻¹):

  1. 227 atm
  2. 2.24 atm
  3. 0.00821 atm
  4. 224 atm
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Answer: (B) 2.24 atm

For a non-electrolyte (i = 1), osmotic pressure π = CRT = 0.1 × 0.0821 × 273 = 2.24 atm.

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12. A 5% (w/v) solution of cane sugar (molar mass = 342 g mol⁻¹) is isotonic with a 1% (w/v) solution of an unknown non-electrolyte substance 'X'. What is the molar mass of substance 'X'?

  1. 136.8 g mol⁻¹
  2. 171.0 g mol⁻¹
  3. 34.2 g mol⁻¹
  4. 68.4 g mol⁻¹
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Answer: (D) 68.4 g mol⁻¹

Since the solutions are isotonic, their molar concentrations are equal: C₁ = C₂. Hence, (w₁ / M₁) = (w₂ / M₂) per 100 mL of solution. This gives 5 / 342 = 1 / Mx, so Mx = 342 / 5 = 68.4 g mol⁻¹.

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