1. M Glucose
2. M Sucrose
3. M Urea
4. M KCl

Solution: M KCl

ΔTf = i Kf m

$\Delta T_{\text{f}}$ is the freezing point depression,

$i$ is the van’t Hoff factor,

$K_{\text{f}}$ is the molal freezing point depression constant for the solvent, and $m$ is the molality of the solution.

In this case, we only need to consider the van’t Hoff factor.

When 1 mol of a solute dissolves, the van’t Hoff factor, i is the number of moles of particles obtained. Sugar and other non-electrolytes do not dissociate in water. One mole of solid sugar yields one mole of sugar molecules when dissolved.

i = 1 for nonelectrolytes

Electrolytes, such as KCl, break down fully into ions.

$\text{KCl}\to \; (K{}^{}+{\text{Cl}}^{\text{–}}$

One mole of solid K$\text{Cl}$ gives two moles of dissolved particles: 1 mol of K${}^{}$ ions and 1 mol of ${\text{Cl}}^{\text{–}}$ ions. Thus, for K$\text{Cl},i=\mathrm{2,\; and\; thus\; the\; depression\; in\; freezing\; point\; will\; be\; maximum\; for\; KCl.}$