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(D) The depression in freezing point is given by the formula $\Delta T_f = i 	imes K_f 	imes m$,where $i$ is the van't Hoff factor and $m$ is the mo

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$0.04 \, m \, Al_2(SO_4)_3$

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(D) The depression in freezing point is given by the formula $\Delta T_f = i 	imes K_f 	imes m$,where $i$ is the van't Hoff factor and $m$ is the molality. The freezing point is lowest when the depression in freezing point $(\Delta T_f)$ is maximum. This depends on the value of $i 	imes m$.For $A$: $KNO_3 ightarrow K^+ + NO_3^-$,so $i = 2$. $i 	imes m = 2 	imes 0.08 = 0.16$.For $B$: $FeCl_3 ightarrow Fe^{3+} + 3Cl^-$,so $i = 4$. $i 	imes m = 4 	imes 0.03 = 0.12$.For $C$: $Ca(NO_3)_2 ightarrow Ca^{2+} + 2NO_3^-$,so $i = 3$. $i 	imes m = 3 	imes 0.05 = 0.15$.For $D$: $Al_2(SO_4)_3 ightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$. $i 	imes m = 5 	imes 0.04 = 0.20$.Since the value of $i 	imes m$ is highest for $D$ $(0.20)$,it will have the maximum depression in freezing point,and thus the lowest freezing point.

Which of the following solutions will have the lowest freezing point?

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