The solution from the following with highest depression in freezing point/lowest freezing point is

We need the solution with highest depression in freezing point. $$\Delta T_f = iK_fm$$ where $$i$$ is van't Hoff factor and $$m$$ is molality.

Option 1: 180 g acetic acid (M=60) in 1 L water. Moles = 3. $$m \approx 3$$. Acetic acid dissociates slightly in water, $$i > 1$$. $$\Delta T_f \approx 3 \times 1.86 \times i > 5.58$$.

Option 2: 180 g acetic acid in benzene. Acetic acid dimerizes in benzene, $$i \approx 0.5$$. $$\Delta T_f$$ much lower.

Option 3: 180 g benzoic acid (M=122) in benzene. Moles $$\approx 1.48$$. Benzoic acid also dimerizes in benzene, $$i \approx 0.5$$.

Option 4: 180 g glucose (M=180) in water. Moles = 1. Non-electrolyte, $$i = 1$$. $$\Delta T_f = 1.86$$.

Option 1 has the highest molality (3 mol/kg) and in water, acetic acid ionizes ($$i > 1$$), giving the largest freezing point depression.

The correct answer is Option 1.