Question

Explain with vapour pressure-temperature curves that the freezing point of a solvent is lowered by dissolving a nonvolatile solute into it. Give reason for such lowering of freezing of solvent.

Answer 1

1. The vapour pressures of solution and of pure solvent are plotted as a function of temperature in the diagram.
   
   Variation of vapour pressure with a temperature of pure solvent, solid solvent and solution
2. The diagram consists of three curves. AB is the vapour pressure curve of solid solvent while CD is the vapour pressure curve of pure liquid solvent. EF is the vapour pressure curve of solution that always lies below the pure solvent.
3. The curves AB and CD intersect at point B where solid and liquid phases of pure solvent are in equilibrium. The two phases have the same vapour pressure at B. The temperature corresponding to B is the freezing point of a solvent `("T"_"f"^0)`.
4. Similarly at E, the point of intersection of EF and AB, the solid solvent and solution are in equilibrium. They have the same vapour pressure at E. The temperature corresponding to E is the freezing point of a solution `("T"_"f")`.
5. It is clear from the diagram that freezing point of solution `("T"_"f")` is lower than that of pure solvent `("T"_"f"^0)` because the vapour pressure curve of solution lies below that of solvent.
6. Reason for lowering of freezing of the solvent:
   At the freezing point of a pure liquid, the attractive forces among molecules are large enough to cause the change of phase from liquid to solid. In a solution, the solvent molecules are separated from each other because of solute molecules. Thus, the separation of solvent molecules in solution is more than that in pure solvent. This results in decreasing the attractive forces between solvent molecules. Consequently, the temperature of the solution is lowered below the freezing point of the solvent to cause the phase change. Hence, freezing point of solvent is lowered by dissolving a nonvolatile solute into it.