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

${\mathrm{P}}_{\mathrm{s}}=\frac{80}{100}{\mathrm{P}}^{\circ };{\mathrm{W}}_{\mathrm{B}}=\mathrm{w}$.t. of solute $=?;\mathrm{mol}$ w.t of solute, ${\mathrm{M}}_{\mathrm{B}}=40 \mathrm{g} {\mathrm{mol}}^{-}$ 

wt. of solvent; ${\mathrm{W}}_{\mathrm{A}}=114 \mathrm{g}$

$\text{g. mol. wt. }\left({\mathrm{M}}_{\mathrm{A}}\right)$ of octane $\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)=(8\times 12)+(18\times 1)$$=114 \mathrm{g} {\mathrm{mol}}^{-}$ 

We know, that 

             $\begin{array}{r}\frac{{\mathrm{P}}^{\circ }-{\mathrm{P}}_{\mathrm{S}}}{{\mathrm{P}}^{\circ }}=\frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}\times \frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{W}}_{\mathrm{A}}}\\ \frac{{\mathrm{P}}^{\circ }-\frac{80}{100}{\mathrm{P}}^{\circ }}{{\mathrm{P}}^{\circ }}=\frac{{\mathrm{W}}_{\mathrm{B}}\times 114}{40\times 114}\\ \end{array}$

${\mathrm{W}}_{\mathrm{B}}=8 \mathrm{g}$