A solution contains 1 mol of pentane $\left(p^{\ast }=450\mathrm{mmHg}\right)$ and 4 mol of hexane $\left(p^{\ast }=150\mathrm{mmHg}\right)$, the mole fraction of pentane in vapour phase will be

In vapour phase, we will have

$\begin{array}{l}{p}_{\text{pentane }}=\{1/(1+4\left)\right\}\left(450\mathrm{mmHg}\right)=90\mathrm{mmHg};p_{\text{pentane }}=\{4/(1+4\left)\right\}\left(150\mathrm{mmHg}\right)=120\mathrm{mmHg}\\ y_{\text{pentane }}=\frac{p_{\text{pentane }}}{p_{\text{pentane }}+p_{\text{hexane }}}=\frac{90}{90+120}=0.429\end{array}$