Use the freezing point depression formula

$\Delta T_f = i K_f m$

where:

• $i$ = van’t Hoff factor
• $K_f$ = cryoscopic constant of the solvent
• $m$ = molality

The solution with largest $iK_fm$ will have the highest depression in freezing point and hence lowest freezing point.

Find moles of each solute from $180\,\text{g}$

• Acetic acid, $\mathrm{CH_3COOH}$, molar mass $= 60\,\text{g mol}^{-1}$ $n = \frac{180}{60} = 3\,\text{mol}$

• Benzoic acid, $\mathrm{C_6H_5COOH}$, molar mass $= 122\,\text{g mol}^{-1}$ $n = \frac{180}{122} \approx 1.48\,\text{mol}$

• Glucose, $\mathrm{C_6H_{12}O_6}$, molar mass $= 180\,\text{g mol}^{-1}$ $n = \frac{180}{180} = 1\,\text{mol}$

Compare van’t Hoff factor $i$ in each solvent

• Acetic acid in benzene: associates (dimerizes) in benzene, so $i<1$; for strong dimerization, approximately $i \approx 0.5$
• Acetic acid in water: weak acid, partially ionizes, so $i>1$ (slightly greater than 1)
• Benzoic acid in benzene: also associates (dimerizes), so $i<1$; approximately $i \approx 0.5$
• Glucose in water: non-electrolyte, so $i=1$

Compare solvent constants $K_f$

Standard values:

• For benzene: $K_f \approx 5.12\,\text{K kg mol}^{-1}$
• For water: $K_f \approx 1.86\,\text{K kg mol}^{-1}$

Compare effective values proportional to $\Delta T_f$

Since masses of solvent are not specified, we compare on the basis of same solvent mass assumption.

Option A: Acetic acid in benzene

$\Delta T_f \propto iK_f n \approx 0.5 \times 5.12 \times 3 = 7.68$

Option B: Acetic acid in water

Here $i$ is only slightly greater than 1; even if we take about $i\approx 1$ to $1.05$: $\Delta T_f \propto iK_f n \approx 1 \times 1.86 \times 3 = 5.58$ or slightly more, but still much less than option A.

Option C: Benzoic acid in benzene

$\Delta T_f \propto 0.5 \times 5.12 \times 1.48 \approx 3.79$

Option D: Glucose in water

$\Delta T_f \propto 1 \times 1.86 \times 1 = 1.86$

Conclusion

The largest depression in freezing point is for: $\boxed{\text{A: }180\,\text{g of acetic acid dissolved in benzene}}$

Hence this solution will have the lowest freezing point.

Compare with stored correct answer

Stored correct answer is B, but our calculation shows A gives greater freezing point depression because benzene has a much larger $K_f$, and even after dimerization acetic acid still gives the largest effective particle contribution.

Therefore, I disagree with the stored answer.