0.5 molal aqueous solution of a weak acid (HX | vedclass

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(B) The depression in freezing point is given by the formula: $\Delta T_{f} = i 	imes K_{f} 	imes m$. For the dissociation of weak acid $HX ightle

Vedclass · Accepted Answer

$1.12$

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(B) The depression in freezing point is given by the formula: $\Delta T_{f} = i 	imes K_{f} 	imes m$. For the dissociation of weak acid $HX ightleftharpoons H^{+} + X^{-}$,the van't Hoff factor $i$ is calculated as $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation. Given $\alpha = 20\% = 0.2$,so $i = 1 + 0.2 = 1.2$. Given molality $m = 0.5\, mol\, kg^{-1}$ and $K_{f} = 1.86\, K\, kg\, mol^{-1}$. Substituting these values: $\Delta T_{f} = 1.2 	imes 1.86 	imes 0.5 = 1.116\, K \approx 1.12\, K$.

$0.5$ molal aqueous solution of a weak acid $(HX)$ is $20\%$ ionised. If $K_f$ for water is $1.86\, K\, kg\, mol^{-1}$,the lowering in freezing point of the solution is $........ K$.

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