JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Solutions

Adding a non-volatile solute lowers a solvent's vapour pressure at every temperature (Raoult's law), so the solution's vapour-pressure curve lies below the pure liquid solvent's curve throughout.

The freezing point of a phase is where its liquid vapour-pressure curve meets the vapour-pressure curve of the frozen (solid) phase. Because the solution curve sits lower than the pure-solvent curve, it crosses the frozen-solid curve at a lower temperature than the pure liquid solvent does: $T_f<T_f^{\circ }$.

The correct graph shows a single frozen-solid curve, the pure liquid-solvent curve crossing it at the higher temperature $T_f^{\circ }$, and the solution curve crossing the same frozen-solid curve at the lower temperature $T_f$, with $\Delta T_f=T_f^{\circ }-T_f$ marked as the gap between these two crossing points.