Which one of the following 0.06 M aqueous solutions has lowest freezing point?

The depression in freezing point is given by $$\Delta T_f = i \, K_f \, m$$, where $$i$$ is the van 't Hoff factor (the number of particles the solute produces in solution), $$K_f$$ is the cryoscopic constant, and $$m$$ is the molality. Since all solutions have the same concentration (0.06 M), the solution with the largest $$i$$ will have the greatest freezing point depression and hence the lowest freezing point.

$$\text{C}_6\text{H}_{12}\text{O}_6$$ (glucose) is a non-electrolyte: $$i = 1$$. $$\text{KI}$$ dissociates into $$\text{K}^+$$ and $$\text{I}^-$$: $$i = 2$$. $$\text{K}_2\text{SO}_4$$ dissociates into $$2\text{K}^+$$ and $$\text{SO}_4^{2-}$$: $$i = 3$$. $$\text{Al}_2(\text{SO}_4)_3$$ dissociates into $$2\text{Al}^{3+}$$ and $$3\text{SO}_4^{2-}$$: $$i = 5$$.

Since $$\text{Al}_2(\text{SO}_4)_3$$ produces the most ions ($$i = 5$$), its 0.06 M solution will have the largest $$\Delta T_f$$ and therefore the lowest freezing point.

The answer is option (1).