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(B) The dissociation of the weak acid is represented as: $HX \rightleftharpoons H^{+} + X^{-}$.For the dissociation of an electrolyte,the van't Hoff f

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$1.12$

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(B) The dissociation of the weak acid is represented as: $HX ightleftharpoons H^{+} + X^{-}$.For the dissociation of an electrolyte,the van't Hoff factor $i$ is given by $i = 1 + \alpha(n-1)$,where $n$ is the number of ions produced per molecule.Here,$n = 2$ and $\alpha = 0.20$ $(20 \%)$.So,$i = 1 + 0.20(2-1) = 1.2$.The lowering in freezing point is given by $\Delta T_{f} = i 	imes K_{f} 	imes m$.Substituting the values: $\Delta T_{f} = 1.2 	imes 1.86 	imes 0.5 = 1.116 \ K \approx 1.12 \ K$.

$0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20 \%$ ionized. If $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$,the lowering in freezing point of the solution is: (in $K$)

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