\(P_s=80 \%\) of \(P^{\circ}\)
\(=\frac{80}{100} P^{\circ}=0.8 P^{\circ}\)
Let \(W g\) of solute is present in mixture.
Moles of solute present \(=\frac{W}{40}\) moles
Molar mass of octane, \(\mathrm{C}_8 \mathrm{H}_{18}\)
\(=8 \times 12+18=114 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\therefore\) Moles of octane \(=\frac{114}{114}=1 \mathrm{~mol}\)
Now, \(\frac{P^{\circ}-P_s}{P^{\circ}}=x_2=\frac{n_2}{n_1+n_2}=\frac{W / 40}{\frac{W}{40}+1}\)
\(\frac{P^{\circ}-0 \cdot 80 P^{\circ}}{P^{\circ}}=\frac{W / 40}{(W / 40)+1}\)
\(\begin{aligned}
1-0.80 &=\frac{W \times 40}{40(W+40)}=\frac{W}{W+40} \\
0 \cdot 20=& \frac{W}{W+40} \\
0.2 W+8 &=W \\
8 &=W(1-0.2) \\
8 &=0.8 W
\end{aligned}\)
\(\therefore \quad W=\frac{8}{0.8}=10 \mathrm{~g} .\)