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(A) For $KCl$,the van't Hoff factor $i = 2$ (as $KCl \rightarrow K^+ + Cl^-$). For $BaCl_2$,the van't Hoff factor $i = 3$ (as $BaCl_2 \rightarrow Ba^{

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$-3 \ ^\circ C$

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(A) For $KCl$,the van't Hoff factor $i = 2$ (as $KCl ightarrow K^+ + Cl^-$). For $BaCl_2$,the van't Hoff factor $i = 3$ (as $BaCl_2 ightarrow Ba^{2+} + 2Cl^-$). Since the concentration is the same,the depression in freezing point $\Delta T_f$ is directly proportional to $i$ $(\Delta T_f = i 	imes K_f 	imes m)$. Therefore,$\frac{\Delta T_f(KCl)}{\Delta T_f(BaCl_2)} = \frac{i(KCl)}{i(BaCl_2)} = \frac{2}{3}$. Given $\Delta T_f(KCl) = 0 - (-2) = 2 \ ^\circ C$. So,$\frac{2}{\Delta T_f(BaCl_2)} = \frac{2}{3}$,which gives $\Delta T_f(BaCl_2) = 3 \ ^\circ C$. The freezing point of the $BaCl_2$ solution is $0 - 3 = -3 \ ^\circ C$.

$A$ $0.01 \ M$ solution of $KCl$ and $BaCl_2$ is prepared in water. If the freezing point of the $KCl$ solution is $-2 \ ^\circ C$,what will be the freezing point of the $BaCl_2$ solution,assuming complete ionization?

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