The vapour pressure of pure liquid X and pure liquid Y at 25 °C are 120 mm Hg and 160 mm Hg respectively. If equal moles of X and Y are mixed to form…

\[\begin{aligned} \mathrm{P}_{\text {Total }} & =\mathrm{p}^{\circ}{ }_{\mathrm{X}} \chi_{\mathrm{X}}+\mathrm{p}^{\circ}{ }_{\mathrm{Y}} \chi_{\mathrm{B}} \\ \chi_{\mathrm{X}} & =\chi_{\mathrm{B}}=0.5 \\ \mathrm{P}_{\text {Total }} & =(120 \times 0.5)+(160 \times 0.5) \\ & =140 \mathrm{~mm} \mathrm{Hg} \end{aligned} \]