If the freezing point of an aqueous solution | vedclass

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(A) For salt $MX_4$,the dissociation is $MX_4 \rightarrow M^{4+} + 4X^-$. The number of ions $n = 5$.Van't Hoff factor $i = 1 + (n-1)\alpha = 1 + (5-1

Vedclass · Accepted Answer

$100.9$

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(A) For salt $MX_4$,the dissociation is $MX_4 ightarrow M^{4+} + 4X^-$. The number of ions $n = 5$.Van't Hoff factor $i = 1 + (n-1)\alpha = 1 + (5-1) 	imes 0.9 = 1 + 3.6 = 4.6$.Freezing point depression $\Delta T_f = T_f^{\circ} - T_f = 0 - (-3) = 3\,K$.Using $\Delta T_f = i 	imes k_f 	imes m$,we get $3 = 4.6 	imes 1.86 	imes m$,so $m = 3 / (4.6 	imes 1.86) \approx 0.3506\,m$.At boiling point,the salt is $100\%$ ionized,so $i' = n = 5$.Boiling point elevation $\Delta T_b = i' 	imes k_b 	imes m = 5 	imes 0.52 	imes 0.3506 \approx 0.911\,K$.Boiling point $T_b = T_b^{\circ} + \Delta T_b = 100 + 0.911 = 100.911\,^{\circ}C$.

If the freezing point of an aqueous solution of salt $MX_4$ $(\alpha = 90\%)$ is $-3\,^{\circ}C$,then its boiling point will be ......... $^{\circ}C$. (Assume it is $100\%$ ionized at boiling temperature and $k_f = 1.86\,K\,kg\,mol^{-1}, k_b = 0.52\,K\,kg\,mol^{-1}$)

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