Q.What is a vapour pressure of liquid? What is relative lowering of vapour pressure?

Step 1. Vapour pressure of a liquid. When a liquid evaporates inside a closed container, evaporation and condensation both proceed until their rates become equal -- a dynamic equilibrium between liquid and vapour. The pressure exerted by the vapour at this equilibrium is the vapour pressure of the liquid at that temperature.

Step 2. Relative lowering of vapour pressure. When a nonvolatile solute is dissolved in the liquid (now the solvent), the resulting solution's vapour pressure $P_{solution}$ is lower than the pure solvent's vapour pressure $P^\circ_{solvent}$. The fractional decrease, $\dfrac{P^\circ_{solvent}-P_{solution}}{P^\circ_{solvent}}$, is called the relative lowering of vapour pressure.

Step 3. By Raoult's law, this relative lowering is numerically equal to the solute's mole fraction, $x_{solute}$ -- making it a colligative property (independent of the solute's chemical identity).

Define vapour pressure via the dynamic liquid-vapour equilibrium, then define relative lowering as the fractional drop in that pressure on dissolving a nonvolatile solute, and link it to the solute's mole fraction via Raoult's law.

• Defining vapour pressure without mentioning the CLOSED container / dynamic equilibrium condition -- an open container never reaches this equilibrium.
• Giving the ABSOLUTE lowering ($P^\circ-P$) instead of the RELATIVE (fractional) lowering the question specifically asks for.