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A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86 °C/m, the freezing point of the solution will be:

1. -0.18 °C
2. -0.54 °C
3. -0.36 °C
4. -0.24 °C

Correct answer: -0.54 °C

Solution

The freezing point depression is calculated using ΔTf = i × Kf × m. Here, m = 0.1 molal, Kf = 1.86 °C/m, and the van't Hoff factor i = 1 + α(n-1), where α = 0.3 and n = 2 (since the weak acid dissociates into 2 particles). Substituting, i = 1 + 0.3(2-1) = 1.3. Thus, ΔTf = 1.3 × 1.86 × 0.1 = 0.2418 °C, approximately -0.24 °C.

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