JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Solutions

Depression of freezing point follows $\Delta T_f=i{K}_{f}m$. Taking the solvent mass as the same in each case, the molalities are $3$ for the two acetic-acid solutions ($180/60$), about $1.48$ for benzoic acid ($180/122$) and $1$ for glucose ($180/180$). In benzene both carboxylic acids dimerise through hydrogen bonding, so their van't Hoff factor is about $\frac{1}{2}$; in water acetic acid is only weakly ionised, so its factor is barely above one, while glucose stays a non-electrolyte with factor one. Using $K_f=1.86$ for water and $5.12$ for benzene, the depressions come out to roughly $5.6$ for acetic acid in water, $7.7$ for acetic acid in benzene, $3.8$ for benzoic acid in benzene and $1.9$ for glucose in water, so the largest depression belongs to acetic acid in benzene. The quantity actually asked for, however, is the lowest freezing temperature, and the two solvents do not start from the same point: water freezes at $0^{\circ }\mathrm{C}$ whereas benzene freezes near $5.5^{\circ }\mathrm{C}$. Subtracting each depression from its own solvent's freezing point gives about $-5.6^{\circ }\mathrm{C}$ for acetic acid in water against only about $-2.2^{\circ }\mathrm{C}$ for acetic acid in benzene, so the aqueous acetic-acid solution reaches the lowest freezing point of the set.