At ${80}^{\circ }\mathrm{C},$, the vapour pressure of pure liquid A is 520 mm$\mathrm{Hg}$ and that of pure liquid B is 1000 mm$\mathrm{Hg}$. If a solution of A and B boils at ${80}^{\circ }\mathrm{C},$ and 1 atm pressure, the amount per cent of A in the mixture is

Let $x_{\mathrm{A}}$  be the amount fraction of A in the liquid solution. We will have

$p_{\text{tocal }}=x_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }+x_{\mathrm{B}}{p}_{\mathrm{B}}^{\ast }=x_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }+\left(1-x_{\mathrm{A}}\right)p_{\mathrm{B}}^{\ast }=x_{\mathrm{A}}\left(p_{\mathrm{A}}^{\ast }-p_{\mathrm{B}}^{\ast }\right)+p_{\mathrm{B}}^{\ast }$This gives $x_{\mathrm{A}}=\frac{p_{\text{total }}-p_{\mathrm{B}}^{\ast }}{p_{\mathrm{A}}^{\ast }-p_{\mathrm{B}}^{\ast }}=\frac{760\mathrm{mmHg}-1000\mathrm{mmHg}}{520\mathrm{mmHg}-1000\mathrm{mmHg}}=\frac{240}{480}=\frac{1}{2}$Hence, the amount per cent of A is $\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}\times 100=\frac{1\mathrm{mol}}{2\mathrm{mol}}\times 100=50$