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(D) The freezing point depression is a colligative property,which depends on the van't Hoff factor $(i)$,representing the number of particles produced

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$1 \ M \ Al_2(SO_4)_3$

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(D) The freezing point depression is a colligative property,which depends on the van't Hoff factor $(i)$,representing the number of particles produced per formula unit in solution.Urea is a non-electrolyte,so $i = 1$.$NaCl$ dissociates as $NaCl ightarrow Na^+ + Cl^-$,so $i = 2$.$Na_2SO_4$ dissociates as $Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-}$,so $i = 3$.$Al_2(SO_4)_3$ dissociates as $Al_2(SO_4)_3 ightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$.Since the freezing point depression $\Delta T_f = i 	imes K_f 	imes m$,the solution with the highest value of $i$ will have the maximum depression in freezing point,resulting in the lowest freezing point.Thus,$1 \ M \ Al_2(SO_4)_3$ has the lowest freezing point. Hence,the correct option is $D$.

Which of the following solutions has the lowest freezing point?

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