The mole fraction of liquid A in a binary liquid solution of $\mathrm{A}\left(p_{\mathrm{A}}^{\ast }=300\text{ Torr }\right)$and $B\left(p_{\mathrm{B}}^{\ast }=800\text{ Torr }\right)$ is 0.6. The pressure at which last droplet of liquid is vapouried and the mole fraction of A in this last droplet, respectively, are

When the last droplet of liquid is vapourized, the composition of vapour formed will be $y_{\mathrm{A}}=0.60$ (given composition of the liquid solution).

Hence, $y_{\mathrm{A}}=\frac{p_{\mathrm{A}}}{p}=\frac{x_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }}{x_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }+x_{\mathrm{B}}{p}_{\mathrm{B}}^{\ast }}=\frac{x_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }}{p_{\mathrm{B}}^{\ast }+\left(p_{\mathrm{A}}^{\ast }-p_{\mathrm{B}}^{\ast }\right)x_{\mathrm{A}}}$

i.e.  $0.60=\frac{x_{\mathrm{A}}\left(300\text{ Torr }\right)}{800\text{ Torr }+(300\text{ Torr }-800\text{ Torr })x_{\mathrm{A}}}$

Solving for $x_{\mathrm{A}}$, we get $x_{\mathrm{A}}=0.80$.

The pressure of the vapour will be $p=x_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }+x_{\mathrm{B}}{p}_{\mathrm{B}}^{\ast }=\left(0.80\right)\left(300\text{ Torr }\right)+\left(0.20\right)\left(800\text{ Torr }\right)=400\text{ Torr }$