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

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$Al_2(SO_4)_3$

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(A) The freezing point depression is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. Since the molality $(m)$ and the solvent $(K_f)$ are the same for all solutions,the freezing point depends on the van't Hoff factor $(i)$. $Al_2(SO_4)_3$ dissociates as $2Al^{3+} + 3SO_4^{2-}$,giving $i = 5$. $C_5H_{10}O_5$ is a non-electrolyte,$i = 1$. $KI$ dissociates as $K^+ + I^-$,giving $i = 2$. $C_{12}H_{22}O_{11}$ is a non-electrolyte,$i = 1$. Since $Al_2(SO_4)_3$ has the highest van't Hoff factor $(i=5)$,it will show the maximum depression in freezing point,resulting in the lowest freezing point.

Which of the following $0.10 \ m$ aqueous solution will have the lowest freezing point?

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