Question

Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL of solution. III. 9.0 g Urea in 500 mL of solution. IV. 20.5 g Aluminium sulphate in 750 mL of solution. The correct increasing order of boiling point of these solutions will be: [Given: Molar mass in g mol\(^{-1}\): H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and S = 32]

• A. I<III<IV<II
• B. III<I<II<IV
• C. I<II<III<IV
• D. III<II<I<IV

Hint:

For colligative properties: - Always calculate effective molality using van’t Hoff factor \(i\). - Strong electrolytes dissociate into multiple ions, increasing particle count and effect.

Answer 1

The Correct Option is B

Concept:
Elevation in boiling point depends on a colligative property and is given by:
ΔTb = i × Kb × m
where i is the van’t Hoff factor and m is the molality.

Step 1: Calculate effective molality for each solution.

I. Glucose (non-electrolyte, i = 1)
Molar mass = 180 g mol-1
Moles = 2.2 / 180 ≈ 0.0122 mol
Mass of solvent = 0.125 kg
Molality = 0.0122 / 0.125 ≈ 0.0976
Effective molality = 0.0976

II. Calcium chloride (CaCl2)
Dissociates into Ca2+ and 2Cl−, so i = 3
Molar mass = 111 g mol-1
Moles = 1.9 / 111 ≈ 0.0171 mol
Mass of solvent = 0.25 kg
Molality = 0.0171 / 0.25 ≈ 0.0684
Effective molality = 3 × 0.0684 ≈ 0.205

III. Urea (non-electrolyte, i = 1)
Molar mass = 60 g mol-1
Moles = 9 / 60 = 0.15 mol
Mass of solvent = 0.5 kg
Molality = 0.15 / 0.5 = 0.30
Effective molality = 0.30

IV. Aluminium sulphate Al2(SO4)3
Dissociates into 2Al3+ and 3SO4 2−, so i = 5
Molar mass = 342 g mol-1
Moles = 20.5 / 342 ≈ 0.0599 mol
Mass of solvent = 0.75 kg
Molality = 0.0599 / 0.75 ≈ 0.0799
Effective molality = 5 × 0.0799 ≈ 0.399

Step 2: Compare effective molalities.
I = 0.0976
II = 0.205
III = 0.30
IV = 0.399

Step 3: Increasing order of boiling point.
I < II < III < IV

Final Answer:
I < II < III < IV