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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol⁻¹, the lowering in freezing point of the solution is

1. 0.56 K
2. 1.12 K
3. 0.56 K
4. 1.12 K

Correct answer: 1.12 K

Solution

The van't Hoff factor (i) is calculated as 1 + α(n-1), where α is the degree of ionization (0.2) and n is the number of particles formed (2 for HX dissociating into H⁺ and X⁻). Thus, i = 1 + 0.2(2-1) = 1.2. The freezing point depression is ΔTf = i × Kf × m = 1.2 × 1.86 × 0.5 = 1.12 K.

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