Q.(i) What is the vapour pressure of pure water at 100°C temperature? (1 mark)

Adding a non-volatile solute reduces the fraction of surface occupied by solvent molecules, lowering the rate of evaporation and hence the equilibrium vapour pressure; Raoult's law quantifies this as a colligative property.

1. Why vapour pressure decreases: In a pure solvent, the entire liquid surface is available for solvent molecules to escape into the vapour phase. When a non-volatile solute is dissolved, some solute particles occupy positions at the surface of the solution, effectively reducing the fraction of the surface covered by solvent molecules. Since fewer solvent molecules are present per unit surface area to escape into the vapour phase, the rate of evaporation of solvent drops. At equilibrium, this gives a lower vapour pressure than that of the pure solvent.
2. Raoult's law (relative lowering of vapour pressure): For a solution of a non-volatile solute in a volatile solvent, the relative lowering of vapour pressure of the solution equals the mole fraction of the solute: $\dfrac{p^0 - p_s}{p^0} = x_{solute}$ …