CLASS 11–12 CHEMISTRY · SOLUTIONS

Boiling point elevation and freezing point depression

ΔTb = i·Kb·m and ΔTf = i·Kf·m, where m is molality. For water Kb = 0.52 and Kf = 1.86 K kg mol⁻¹; i = 1 for non-electrolytes.

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Worked examples

Example 1 · level 1

An aqueous solution of glucose (a non-electrolyte) has molality 0.5 mol/kg. By how many kelvin is its freezing point lowered compared with pure water? Give the answer correct to 2 decimal places. For water, Kb = 0.52 K kg mol⁻¹ and Kf = 1.86 K kg mol⁻¹.

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ΔTf = Kf × m = 1.86 × 0.5 = 0.93 ≈ 0.93 K.

Answer: 0.93 K

Example 2 · level 2

Meera dissolves 12 g of urea (CO(NH₂)₂, a non-electrolyte) in 250 g of water. At what temperature (in °C) will the solution freeze, correct to 2 decimal places? Pure water freezes at 0 °C. For water, Kb = 0.52 K kg mol⁻¹ and Kf = 1.86 K kg mol⁻¹. (Atomic masses: C = 12, O = 16, N = 14, H = 1)

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M(CO(NH₂)₂) = 12 + 16 + 2 × 14 + 4 × 1 = 60 g/mol; n = 12 ÷ 60 = 0.2 mol.
m = 0.2 ÷ 0.25 kg = 0.8 mol/kg.
ΔTf = 1.86 × 0.8 = 1.488 K.
Freezing point = 0 − 1.488 ≈ −1.49 °C.

Answer: -1.49

Example 3 · level 3

0.2 mol of sodium sulphate (Na₂SO₄) is dissolved in 200 g of water. Assuming complete dissociation, find the depression in freezing point in kelvin, correct to 2 decimal places. For water, Kb = 0.52 K kg mol⁻¹ and Kf = 1.86 K kg mol⁻¹.

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Na₂SO₄ → 2Na⁺ + SO₄²⁻, so i = 3.
m = 0.2 ÷ 0.2 = 1 mol/kg.
ΔTf = 3 × 1.86 × 1 = 5.58 ≈ 5.58 K.

Answer: 5.58 K