What weight of non-volatile solute of molar mass 40 , should be dissolved in 57 gr of octane to reduce it's vapour pressure to 80 %

The vapour pressure of pure octane =${\mathrm{P}}^0$
The vapour pressure of octane after dissolving the non-volatile solute is
$\left({\mathrm{P}}^{\mathrm{s}}\right)=\frac{80}{100}=0.8$
Molar mass of solute$\left({\mathrm{M}}_2\right)=40\mathrm{gr}$
Mass of octane $\left({\mathrm{W}}_1\right)=57$
Molar mass of octane $\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right){\mathrm{M}}_1=114$
$$\begin{gathered} \frac{{\mathrm{P}}^0-{\mathrm{P}}^{\mathrm{S}}}{{\mathrm{P}}^0}=\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}\times \frac{\mathrm{M}1}{{\mathrm{W}}_1}\Rightarrow \frac{{\mathrm{P}}^0-0.8}{{\mathrm{P}}^0}=\frac{{\mathrm{W}}_2\times 114}{40\times 57} \\ 0.2=\frac{{\mathrm{W}}_2\times 2}{40}\Rightarrow {\mathrm{W}}_2=\frac{0.2\times 40}{2}=4 \\ {\mathrm{W}}_2=4 \end{gathered}$$