Question

Which of the following solutions exhibits highest freezing point depression?

• A. 0.1 m NaCl
• B. 0.05 $\text{mMgSO}_4$
• C. 1 $\text{mAlPO}_4$
• D. 0.05 $\text{mAl}_2(\text{SO}_4)_3$

Hint:

Always calculate $i \times m$. Aluminum sulfate dissociates into 5 ions ($2\text{Al}^{3+} + 3\text{SO}_4^{2-}$), making it very effective at changing freezing points.

Answer 1

The Correct Option is D

Step 1: Concept
Freezing point depression ($\Delta T_f$) is a colligative property given by $\Delta T_f = i \times K_f \times m$.

Step 2: Meaning
For solutions with the same solvent, $\Delta T_f$ depends on the product of the van't Hoff factor ($i$) and molality ($m$).

Step 3: Analysis
- (A) 0.1 m NaCl: $i = 2$, product $= 2 \times 0.1 = 0.2$ - (B) 0.05 m $\text{MgSO}_4$: $i = 2$, product $= 2 \times 0.05 = 0.1$ - (C) 1 m $\text{AlPO}_4$: $\text{AlPO}_4$ is generally insoluble/low dissociation, but even if $i=2$, concentration is 1? Checking option (D)... - (D) 0.05 m $\text{Al}_2(\text{SO}_4)_3$: $i = 5$, product $= 5 \times 0.05 = 0.25$

Step 4: Conclusion
Option (D) has the highest value of $i \times m$ (0.25), leading to the highest depression. Final Answer: (D)