For a 0.1 , molal aqueous solution of 2-iodop | vedclass

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(B) The dissociation of $2-$iodopropanoic acid $(CH_3CH(I)COOH)$ is given by: $CH_3CH(I)COOH \rightleftharpoons CH_3CH(I)COO^- + H^+$. For a weak acid

Vedclass · Accepted Answer

$-0.1953$

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(B) The dissociation of $2-$iodopropanoic acid $(CH_3CH(I)COOH)$ is given by: $CH_3CH(I)COOH ightleftharpoons CH_3CH(I)COO^- + H^+$. For a weak acid,the van't Hoff factor $i = 1 + \alpha(n-1)$. Here,$n = 2$ and $\alpha = 0.05$ $(5 \%)$. So,$i = 1 + 0.05(2-1) = 1.05$. The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Given $K_f = 1.86 \, K \, kg \, mol^{-1}$ and $m = 0.1 \, molal$. $\Delta T_f = 1.05 	imes 1.86 	imes 0.1 = 0.1953 \, K$. The freezing point of the solution is $T_f = T_f^0 - \Delta T_f = 0 - 0.1953 = -0.1953 \, ^oC$.

For a $0.1 \, molal$ aqueous solution of $2-$iodopropanoic acid,the degree of dissociation is $5 \%$. The freezing point of the solution will be ............ $^o C$.

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