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easyNEET 2021

1. Elevation of boiling point is a _________ property:

  1. Additive
  2. Colligative
  3. Constitutive
  4. Intensive
Show answer and explanation

Answer: (B) Colligative

Colligative properties depend on number of solute particles, not their nature: ΔTb, ΔTf, osmotic pressure, relative lowering of VP.

Quick tip: 4 colligative properties: ΔTb, ΔTf, osmotic pressure, relative lowering of VP. All depend on number of particles.

NCERT Class 12, Chapter 2, page 44

easy

2. Acetic acid associates as dimers in benzene. If the degree of association of acetic acid in benzene is 0.8, the van't Hoff factor (i) is:

  1. 0.6
  2. 0.4
  3. 0.2
  4. 0.8
Show answer and explanation

Answer: (A) 0.6

For dimerization (2CH₃COOH ⇌ (CH₃COOH)₂), n = 2. The formula for the van't Hoff factor is i = 1 - α(1 - 1/n) = 1 - 0.8(1 - 1/2) = 1 - 0.4 = 0.6.

medium

3. Arrange the following 0.10 m aqueous solutions in order of increasing boiling point: I. 0.10 m Al₂(SO₄)₃ (α = 0.8) II. 0.10 m BaCl₂ (α = 0.9) III. 0.10 m Urea IV. 0.10 m NaCl (α = 0.9)

  1. IV < III < II < I
  2. III < II < IV < I
  3. III < IV < II < I
  4. I < II < IV < III
Show answer and explanation

Answer: (C) III < IV < II < I

ΔTb is directly proportional to i × m. For Urea: i = 1, i × m = 0.10. For NaCl: i = 1 + 0.9 = 1.9, i × m = 0.19. For BaCl₂: n = 3, i = 1 + 2(0.9) = 2.8, i × m = 0.28. For Al₂(SO₄)₃: n = 5, i = 1 + 4(0.8) = 4.2, i × m = 0.42. Higher i × m means higher boiling point. Thus, the order of increasing boiling point is III < IV < II < I.

hard

4. Two liquids X and Y form an ideal solution. At a given temperature, the vapour pressure of pure X is 400 mm Hg and that of pure Y is 600 mm Hg. If the mole fraction of X in the vapour phase is 0.50, what is the mole fraction of X in the liquid phase?

  1. 0.40
  2. 0.50
  3. 0.75
  4. 0.60
Show answer and explanation

Answer: (D) 0.60

In the vapour phase, y_X = 0.50, which means partial pressures are equal: p_X = p_Y. Using Raoult's law: p_X° × x_X = p_Y° × (1 - x_X) ⇒ 400 × x_X = 600 × (1 - x_X) ⇒ 1000 x_X = 600 ⇒ x_X = 0.60.

medium

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

  1. 0.744 K
  2. 0.186 K
  3. 9.3 K
  4. 0.372 K
Show answer and explanation

Answer: (D) 0.372 K

ΔT_f = i K_f m = 1 × 1.86 × 0.2 = 0.372 K (van't Hoff factor i = 1).

easy

6. What is the relationship between the ebullioscopic constant (Kb) and cryoscopic constant (Kf) for a solvent with gas constant R, molar mass M, normal boiling point Tb, normal freezing point Tf, enthalpy of vaporization ΔvapH, and enthalpy of fusion ΔfusH?

  1. Kf = (R · M · Tf) / (1000 · ΔfusH)
  2. Kb = (1000 · R · Tb²) / (M · ΔvapH)
  3. Kb = (R · M · Tb²) / (1000 · ΔvapH)
  4. Kf = (1000 · R · Tf²) / (M · ΔfusH)
Show answer and explanation

Answer: (C) Kb = (R · M · Tb²) / (1000 · ΔvapH)

According to thermodynamic relations given in NCERT, the molal elevation constant is Kb = (R · M · Tb²) / (1000 · ΔvapH) and the molal depression constant is Kf = (R · M · Tf²) / (1000 · ΔfusH), where M is the molar mass of the solvent.

hard

7. At 300 K, pure benzene (molar mass = 78 g mol⁻¹) and pure toluene (molar mass = 92 g mol⁻¹) have vapour pressures of 75 Torr and 25 Torr, respectively. In an ideal solution containing equal masses of both components, what is the mole fraction of benzene in the vapour phase?

  1. 0.78
  2. 0.65
  3. 0.54
  4. 0.86
Show answer and explanation

Answer: (A) 0.78

Let the mass of each be m. Moles of benzene = m/78 and toluene = m/92. Mole fraction in liquid: x_B = (m/78) / (m/78 + m/92) = 92 / 170 = 0.541. Partial pressures: P_B = 75 × 0.541 = 40.59 Torr; P_T = 25 × (1 - 0.541) = 11.47 Torr. Total pressure = 52.06 Torr. In vapour phase, y_B = P_B / P_total = 40.59 / 52.06 ≈ 0.78.

easy

8. 29.3 g of NaCl (M = 58.5 g/mol) is dissolved in 250 g of water. The molality of the solution is:

  1. 2 mol/kg
  2. 0.125 mol/kg
  3. 0.117 mol/kg
  4. 0.002 mol/kg
Show answer and explanation

Answer: (A) 2 mol/kg

Moles = 29.3/58.5 = 0.5. Molality = 0.5/(250/1000 kg) = 2 mol/kg.

easy

9. If a salt XY₃ dissociates in water with a degree of dissociation α, which of the following expressions correctly relates the van’t Hoff factor (i) to α?

  1. i = 1 + 3α
  2. i = 1 + 2α
  3. i = 1 + 4α
  4. i = 1 - 3α
Show answer and explanation

Answer: (A) i = 1 + 3α

For the dissociation XY₃ ⇌ X³⁺ + 3Y⁻, the number of ions produced per formula unit is n = 4. The relationship between i and α is i = 1 + (n - 1)α = 1 + (4 - 1)α = 1 + 3α.

medium

10. A solution containing 18 g of a non-volatile non-electrolyte solute in 200 g of water freezes at -0.93 °C. What is the molar mass of the solute? (Kf for water = 1.86 K kg mol⁻¹)

  1. 180 g mol⁻¹
  2. 90 g mol⁻¹
  3. 60 g mol⁻¹
  4. 342 g mol⁻¹
Show answer and explanation

Answer: (A) 180 g mol⁻¹

ΔTf = Kf × (w₂ × 1000) / (M₂ × w₁). Substituting values: 0.93 = 1.86 × (18 × 1000) / (M₂ × 200). Thus, 0.93 = 1.86 × 90 / M₂, which gives M₂ = (1.86 × 90) / 0.93 = 180 g mol⁻¹.

medium

11. Given below are two statements: Statement I: The value of van’t Hoff factor (i) for an electrolyte that undergoes complete ionization into two ions is equal to 2. Statement II: For a non-volatile solute undergoing association to form pentamers (5A ⇌ A₅) with 100% efficiency, the value of i is 0.20. In the light of the above statements, choose the most appropriate answer from the options given below:

  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
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

For complete dissociation into two ions (like NaCl), i = 1 + (2 - 1)(1) = 2. For complete pentamerization (5A ⇌ A₅), 5 molecules aggregate into 1, so i = 1/5 = 0.20. Therefore, both Statement I and Statement II are correct.

medium

12. For an ideal solution composed of two volatile liquids A and B, if P_A° = 300 mm Hg and P_B° = 100 mm Hg, at what mole fraction of A in the liquid phase will the vapour contain 75 mol % of A?

  1. 0.50
  2. 0.25
  3. 0.75
  4. 0.60
Show answer and explanation

Answer: (A) 0.50

The relationship is y_A = (x_A · P_A°) / [x_A · P_A° + (1 - x_A) · P_B°]. Substituting values: 0.75 = (300 x_A) / [300 x_A + 100(1 - x_A)] = 300 x_A / [200 x_A + 100]. This gives 0.75(200 x_A + 100) = 300 x_A → 150 x_A + 75 = 300 x_A → 150 x_A = 75 → x_A = 0.50.

medium

13. A mixture of ethanol and acetone shows positive deviation from Raoult's law because:

  1. Strong hydrogen bonds form between them
  2. They react chemically
  3. Their volumes decrease on mixing
  4. Mixing breaks hydrogen bonds in ethanol, so A-B interactions are weaker
Show answer and explanation

Answer: (D) Mixing breaks hydrogen bonds in ethanol, so A-B interactions are weaker

Acetone molecules come between ethanol molecules and break some hydrogen bonds. Weaker A-B interactions mean higher vapour pressure than predicted (positive deviation).

NCERT Class 12

easy

14. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: A 0.1 M aqueous solution of potassium chloride has a higher osmotic pressure than a 0.1 M aqueous solution of glucose at 298 K. Reason R: Potassium chloride undergoes ionization into ions in water, whereas glucose remains un-ionized as neutral molecules. In light of the above statements, choose the most appropriate answer:

  1. Both Assertion A and Reason R are true and Reason R is the correct explanation of Assertion A
  2. Both Assertion A and Reason R are true but Reason R is not the correct explanation of Assertion A
  3. Assertion A is true but Reason R is false
  4. Assertion A is false but Reason R is true
Show answer and explanation

Answer: (A) Both Assertion A and Reason R are true and Reason R is the correct explanation of Assertion A

Osmotic pressure depends on the total concentration of dissolved solute particles. KCl dissociates into K⁺ and Cl⁻ (i ≈ 2), giving approximately double the number of particles compared to glucose (i = 1), resulting in a higher osmotic pressure. Reason R correctly explains Assertion A.

easy

15. A solution has concentration 100 mmol/L. Express this in mol/L:

  1. 0.1 mol/L
  2. 1 mol/L
  3. 100 mol/L
  4. 100000 mol/L
Show answer and explanation

Answer: (A) 0.1 mol/L

Unit check: 1 mmol = 10⁻³ mol, so 100 mmol/L = 0.1 mol/L.

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