The vapour pressure of two pure liquids A and B that form an ideal...

V.P. of solution at $t^{{}^{\circ}} \mathrm{C}=760 \mathrm{ mm}$ [at b.p., V.P. of solution =atompheric pressure] Thus =\mathrm{P}_{\mathrm{A}^{{}^{\circ}} \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}^{\infty} \mathrm{x}_{\mathrm{B}} or \mathrm{P}=\mathrm{P}_{\mathrm{A}^{0} \cdot \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}^{0}\left(1- \mathrm{x}_{\mathrm{A}}\right)\left[\because \mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}=1\right]$or$760=400 \mathrm{X}_{\m mathrm{A}}$+800\lleft(1-$\mathrm{X}_{\mathrm{A}})[\because\mathrm{P}=760 \mathrm{mmofHg}$]$ or $-800+760=-400 x_{\mathrm{A}}$ or $-40=-400$\mathrm{x}_{\mathrm{A}} or x_{\ mmathrm{A}}=\frac{40}{400}=$0.1$ Thus mole fraction in solution is 0.1