$0.2$ molal acid $H X$ is $20$ % ionised in solution. $K_{f} = 1.86 K m o l a l i t y^{- 1}$ . The freezing point of the solution is:

- {0.45}^{\circ} C

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- {0.90}^{\circ} C

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- {0.31}^{\circ} C

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- {0.53}^{\circ} C

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Solution

The correct option is A $- {0.45}^{\circ} C$
$H X \to H^{+} + X^{-}$
initially $1$ $0$ $0$
at eqm $(1 - 0.2) 0.2$ $0.04$ $0.04$
Total moles $= 0.24$
$△ T_{f} = K_{f} m$
$= 1.86 \times 0.24$
$= 0.45$
$f_{p} = 0 - △ T_{f}$
$= - 0.45$

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