Use freezing point depression relation

The depression in freezing point is given by $\Delta T_f = iK_f m$ where:

• $i$ = van’t Hoff factor
• $K_f$ = cryoscopic constant of water
• $m$ = molality

Since all given solutions are dilute aqueous solutions of the same concentration ($0.06\,M$), the one with the largest $i$ will produce the largest depression in freezing point, hence the lowest freezing point.

Find van’t Hoff factor for each solute

(A) $Al_2(SO_4)_3$ $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ Total ions produced $= 2+3=5$ So, $i \approx 5$

(B) $C_6H_{12}O_6$ Glucose is a non-electrolyte and does not dissociate. $i = 1$

(C) KI $KI \rightarrow K^+ + I^-$ Total ions produced $=2$ So, $i \approx 2$

(D) $K_2SO_4$ $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$ Total ions produced $=3$ So, $i \approx 3$

Compare freezing point depression

Since $\Delta T_f \propto i$ the order of depression in freezing point is $Al_2(SO_4)_3 > K_2SO_4 > KI > C_6H_{12}O_6$

Therefore, the solution with the lowest freezing point is: $\boxed{Al_2(SO_4)_3}$

Check with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.