To determine which solution exhibits the largest freezing point depression, we use the formula for freezing point depression:$\\ \Delta T_f = i \cdot K_f \cdot m \\$where:$\\$• $\Delta T_f$ is the freezing point depression.$\\$• $i$ is the van 't Hoff factor, which represents the number of particles the solute splits into.$\\$• $K_f$ is the cryoscopic constant (a property of the solvent).$\\$• $m$ is the molality of the solution.$\\$Since all solutions have the same molality (0.10 m), the largest freezing point depression will occur for the solution with the highest $i. \\$Let's calculate $i$ for each option:$\\$Option 1: $$KCl dissociates into $K^+$ and $Cl^-$ ions.$\\ i = 2 \\$Option 2: $C_6H_{12}O_6$ (glucose) does not dissociate in solution.$\\ i = 1 \\$Option 3: $Al_2(SO_4)_3$ dissociates into $2Al^{3+}$ and $3SO_4^{2-}$ ions.$\\ i = 2 + 3 = 5 \\$Option 4: $K_2SO_4$ dissociates into $2K^+$ and $SO_4^{2-}$ ions.$\\ i = 2 + 1 = 3 \\$The solution with the largest $i$ will have the largest freezing point depression.$\\$Therefore, the solution of $Al_2(SO_4)_3$ (Option 3) will exhibit the largest freezing point depression.$\\$This corresponds to the correct option: 3.