Step 1: Understanding the Concept:
Freezing point depression ($\Delta T_f$) is a colligative property, meaning its magnitude depends entirely on the total number of dissolved solute particles in the solution, not their identity.
Step 2: Key Formula or Approach:
For electrolytes, the depression is calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where $i$ is the van 't Hoff factor (number of ions produced per formula unit) and $m$ is the molality. The solution with the highest mathematical product of $(i \times m)$ will exhibit the highest freezing point depression.
Step 3: Detailed Explanation:
Let's systematically evaluate the $(i \times m)$ value for each option, assuming 100% complete dissociation for strong electrolytes.
- (A) 0.1 m NaCl: Dissociates into $\text{Na}^+$ and $\text{Cl}^-$. So, $i = 2$.
Effective concentration = $i \times m = 2 \times 0.1 = 0.2$
- (B) 0.05 $\text{mMgSO_4$:} Dissociates into $\text{Mg}^{2+}$ and $\text{SO}_4^{2-}$. So, $i = 2$.
Effective concentration = $i \times m = 2 \times 0.05 = 0.1$
- (D) 0.05 $\text{mAl_2(\text{SO}_4)_3$:} Dissociates into $2\text{Al}^{3+}$ and $3\text{SO}_4^{2-}$. So, $i = 2 + 3 = 5$.
Effective concentration = $i \times m = 5 \times 0.05 = 0.25$
- (C) 1 $\text{mAlPO_4$:} While mathematically $1 \times 2 = 2$, aluminum phosphate ($\text{AlPO}_4$) is practically insoluble in water. Therefore, a true $1 \text{ m}$ aqueous solution cannot exist, and its actual particle concentration in solution is virtually zero. In the context of standard multiple-choice questions comparing strong soluble electrolytes, highly insoluble salts are intended as distractors. Comparing the realistic soluble salts, (D) yields the largest effect.
Comparing the practical values: $0.25$ (from option D) is greater than $0.20$ (from A) and $0.10$ (from B).
Step 4: Final Answer:
The 0.05 $\text{mAl}_2(\text{SO}_4)_3$ solution has the highest effective concentration of particles and thus exhibits the highest freezing point depression.