An ideal liquid solution of 1 mol of A and 3 mol of B has the vapour pressure of 550 Torr. If one more mole of B is added, its vapours pressure becomes 560 mmHg. The vapour pressures of pure A and pure B, respectively are

Mole fractions are $x_{\mathrm{A}}=\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}=\frac{1\mathrm{mol}}{(1+3)\mathrm{mol}}=\frac{1}{4}$ and $x_{\mathrm{B}}=1-\frac{1}{4}=\frac{3}{4}$

Hence, $550\text{ Torr }=\frac{1}{4}{p}_{\mathrm{A}}^{\ast }+\frac{3}{4}{p}_{\mathrm{B}}^{\ast }$

When one more mole of B is added, then

$x_{\mathrm{A}}^{\mathrm{\prime }}=\frac{1\mathrm{mol}}{(1+4)\mathrm{mol}}=\frac{1}{5}$ and $x_{\mathrm{B}}^{\mathrm{\prime }}=1-\frac{1}{5}=\frac{4}{5}$

Hence,  $560\mathrm{mmHg}=\frac{1}{5}{p}_{\mathrm{A}}^{\ast }+\frac{4}{5}{p}_{\mathrm{B}}^{\ast }$

Solving $p_{\mathrm{A}}^{\ast }$ and $p_{\mathrm{B}}^{\ast }$ from Eqs (1) and (2), we get $p_{\mathrm{A}}^{\ast }=400\text{ Torr }$and $p_{\mathrm{B}}^{\ast }=600\text{ Torr}$