A solution of CaCl_2 is prepared by dissolvin | vedclass

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(A) $1$. Calculate the molar mass of $CaCl_2$: $M = 40 + 2 	imes 35.5 = 111 \ g \ mol^{-1}$.$2$. Calculate the number of moles of $CaCl_2$: $n = \fra

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$0.0006$

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(A) $1$. Calculate the molar mass of $CaCl_2$: $M = 40 + 2 	imes 35.5 = 111 \ g \ mol^{-1}$.$2$. Calculate the number of moles of $CaCl_2$: $n = \frac{0.0112 \ g}{111 \ g \ mol^{-1}} \approx 1.009 	imes 10^{-4} \ mol$.$3$. Since $CaCl_2$ dissociates as $CaCl_2 ightarrow Ca^{2+} + 2Cl^-$,the van't Hoff factor $(i)$ is $3$.$4$. The molality $(m)$ is $\frac{1.009 	imes 10^{-4} \ mol}{1 \ kg} = 1.009 	imes 10^{-4} \ m$.$5$. The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$.$6$. $\Delta T_f = 3 	imes 2 	imes 1.009 	imes 10^{-4} \approx 6.054 	imes 10^{-4} \approx 0.0006 \ K$.

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