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(A) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Since the concentration $(m)$ and solvent $(K_f)$ are the same

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$-3$

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(A) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Since the concentration $(m)$ and solvent $(K_f)$ are the same for both solutions,$\Delta T_f \propto i$. For $KCl$,$i = 2$ $(KCl ightarrow K^+ + Cl^-)$. For $BaCl_2$,$i = 3$ $(BaCl_2 ightarrow Ba^{2+} + 2Cl^-)$. Given $\Delta T_f$ for $KCl = 0 - (-2) = 2 \ ^oC$. Using the ratio: $\frac{\Delta T_f(KCl)}{\Delta T_f(BaCl_2)} = \frac{i(KCl)}{i(BaCl_2)} = \frac{2}{3}$. $\Delta T_f(BaCl_2) = \frac{3 	imes 2}{2} = 3 \ ^oC$. Freezing point of $BaCl_2 = 0 - 3 = -3 \ ^oC$.

$0.01 \ M$ solutions of $KCl$ and $BaCl_2$ are prepared in water. The freezing point of $KCl$ is found to be $-2 \ ^oC$. What is the freezing point of $BaCl_2$ if it is completely ionised? ............. $^oC$

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