Calculate the mass of a non-volatile solute (molar mass 40g mol^-1) which should be dissolved in 114gm octane to reduce it's vapour pressure to 80%.

Let the vapour pressure of pure octane = 100 mm
$\therefore$ vapour pressure of solution ${\mathrm{p}}^{\mathrm{S}}$ = 80mm
solute: w = ? m = 40
solvent(octane) :
W = 114gm M = 114
$\begin{array}{l}\therefore \frac{{\mathrm{p}}^0-{\mathrm{p}}^{\mathrm{s}}}{{\mathrm{p}}^0}=\frac{\mathrm{w}}{\mathrm{m}}\times \frac{\mathrm{M}}{\mathrm{W}}\\ \Rightarrow \frac{100-80}{100}=\frac{\mathrm{w}}{40}\times \frac{114}{114}\\ \frac{20}{100}=\frac{\mathrm{w}}{40}\\ \therefore \mathrm{w}=8\mathrm{g}\end{array}$