The depression in freezing point ($\Delta T_f$) is a colligative property proportional to the number of solute particles in the solution. It is calculated using the formula $\Delta T_f = i \times K_f \times m$. Since the molality ($m = 0.10 \text{ m}$) and the solvent ($K_f$) are the same for all options, the solution with the highest Van't Hoff factor ($i$) will show the largest freezing point depression.

1. KCl: Strong electrolyte, dissociates as $K^+ + Cl^-$. $i = 2$.
2. C₆H₁₂O₆ (Glucose): Non-electrolyte, does not dissociate. $i = 1$.
3. Al₂(SO₄)₃: Strong electrolyte, dissociates as $2Al^{3+} + 3SO_4^{2-}$. Total ions = 5, so $i = 5$.
4. K₂SO₄: Strong electrolyte, dissociates as $2K^+ + SO_4^{2-}$. Total ions = 3, so $i = 3$.

Since $Al_2(SO_4)_3$ produces the maximum number of particles ($i=5$), it exhibits the largest freezing point depression .