Between aqueous solutions of 0.01 m KCl and | vedclass

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(A) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. For $KCl$,$i = 2$ $(KCl ightarrow K^+ + Cl^-)$. For $BaCl_2$,

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$-3$

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(A) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. For $KCl$,$i = 2$ $(KCl ightarrow K^+ + Cl^-)$. For $BaCl_2$,$i = 3$ $(BaCl_2 ightarrow Ba^{2+} + 2Cl^-)$. Given $\Delta T_{f(KCl)} = 0 - (-2) = 2^{\circ}C$. Since $m$ and $K_f$ are the same for both solutions,we have: $\frac{\Delta T_{f(KCl)}}{i_{KCl}} = \frac{\Delta T_{f(BaCl_2)}}{i_{BaCl_2}}$ $\frac{2}{2} = \frac{\Delta T_{f(BaCl_2)}}{3}$ $\Delta T_{f(BaCl_2)} = 3^{\circ}C$. Therefore,the freezing point of $BaCl_2$ solution = $0 - 3 = -3^{\circ}C$.

Between aqueous solutions of $0.01 \ m \ KCl$ and $0.01 \ m \ BaCl_2$ (strong electrolytes),the freezing point of the $KCl$ solution is $-2^{\circ}C$. The freezing point of the $BaCl_2$ solution will be ..... $^{\circ}C$.

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