Calculate the mass of a non-volatile solute (molar mass $40 \, g \, mol^{-1}$) which should be dissolved in $114 \, g$ octane to reduce its vapour pressure to $80 \%$.

(8 G) Let the vapour pressure of pure octane be $p_{1}^{o}$.
The final vapour pressure is $p_{1} = 0.8 p_{1}^{o}$.
Molar mass of solute,$M_{2} = 40 \, g \, mol^{-1}$.
Mass of octane,$w_{1} = 114 \, g$.
Molar mass of octane $(C_{8}H_{18})$,$M_{1} = (8 \times 12) + (18 \times 1) = 114 \, g \, mol^{-1}$.
Applying Raoult's Law for relative lowering of vapour pressure:
$\frac{p_{1}^{o} - p_{1}}{p_{1}^{o}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$
Substituting the values:
$\frac{p_{1}^{o} - 0.8 p_{1}^{o}}{p_{1}^{o}} = \frac{w_{2} \times 114}{40 \times 114}$
$\frac{0.2 p_{1}^{o}}{p_{1}^{o}} = \frac{w_{2}}{40}$
$0.2 = \frac{w_{2}}{40}$
$w_{2} = 0.2 \times 40 = 8 \, g$.
Hence,the required mass of the solute is $8 \, g$.