Define vapour pressure of a liquid. What happens to the vapour pressure when (a) a volatile solute dissolves in the liquid and (b) non-volatile solute dissolvedin it?

Answer:

Liquids at a given temperature vapourise and under
equilibrium conditions the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure.

When a volatile solute is dissolved into, solvent then the vapour state solute and the solvent.
total vapour pressure above such a solution will be equal to the sum of the pressure exerted by the vapours of both solute and solvent.

${\mathrm{P}}_{\mathrm{Total}} = {\mathrm{p}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}} + {\mathrm{p}}_{\mathrm{B}}^0 {\mathrm{x}}_{\mathrm{B}}$
Where 
P_total = total pressure of solution
$$\begin{gathered} {\mathrm{p}}_{\mathrm{A}}^0 = \mathrm{Vapour} \mathrm{pressure} \mathrm{of} \mathrm{pure} \mathrm{component} \mathrm{A} \\ {\mathrm{p}}_{\mathrm{B}}^0 = \mathrm{Vapour} \mathrm{pressure} \mathrm{of} \mathrm{pure} \mathrm{component} \mathrm{B} \\ {\mathrm{X}}_{\mathrm{A} } = \mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{component} \mathrm{A} \\ {\mathrm{X}}_{\mathrm{B}} = \mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{pure} \mathrm{component} \mathrm{B} \end{gathered}$$

(b)
When  a non- volatile solute is dissovled then there is lowering in vapour pressure .
The lowering in vapour pressureisgiven by $$\begin{gathered} \mathrm{p}° -\mathrm{p} = \mathrm{p}°{\mathrm{x}}_{\mathrm{B}} \\ \mathrm{p} = \mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{solution} \mathrm{containing} \mathrm{non}- \mathrm{volatile} \mathrm{solute} \\ \mathrm{p}° = \mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{pure} \mathrm{solvent}. \end{gathered}$$