A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. The freezing point of this solution is (Given $K_{f} = 1 . 86^{o} C$ for water)

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$- {0.31}^{o} C$

$- {0.45}^{o} C$

$- {0.53}^{o} C$

$- {0.90}^{o} C$

answer is B.

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Detailed Solution

$\begin{array}{l} {ΔT}_{f} = molality \times K_{f} \times (1 + α) \\ α = 0 .2, Molality =0.2, K_{f} = 1 .86 \\ {ΔT}_{f} = 0 .2 \times 1 .2 \times 1 .86 = 0 {.4464}^{o} \\ Freezing point {= -0.45}^{o} C \end{array}$

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