State Raoult's law for a solution containing a non-volatile solute and a volatile solvent and explain it.

(N/A) Raoult's law states that for a solution of volatile liquids,the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in the solution.
For a solution containing a non-volatile solute,only the solvent contributes to the vapour pressure. The law is expressed as: $p_{1} = x_{1} \cdot p_{1}^{0}$,where $p_{1}$ is the vapour pressure of the solvent in the solution,$x_{1}$ is its mole fraction,and $p_{1}^{0}$ is the vapour pressure of the pure solvent.
The reduction in vapour pressure $(\Delta p_{1})$ is given by: $\Delta p_{1} = p_{1}^{0} - p_{1} = p_{1}^{0} - p_{1}^{0} x_{1} = p_{1}^{0}(1 - x_{1})$.
Since $1 - x_{1} = x_{2}$ (where $x_{2}$ is the mole fraction of the solute),we have $\Delta p_{1} = x_{2} \cdot p_{1}^{0}$.
The relative lowering of vapour pressure is given by: $\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}}$.
For dilute solutions,$n_{2} \ll n_{1}$,so $\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} \approx \frac{n_{2}}{n_{1}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$,where $w$ is mass and $M$ is molar mass.