If Two liquids A and B which form a non-ideal solution which obey the  below equation as ${\mathrm{P}}_{\mathrm{T}}={\mathrm{P}}_{\mathrm{A}}^0+3({\mathrm{P}}_{\mathrm{B}}^0-{\mathrm{P}}_{\mathrm{A}}^0){\mathrm{X}}_{\mathrm{B}}+2({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0){\mathrm{X}}_{\mathrm{B}}^2\mathrm{.}$

Calculate the exact composition (by mole) when this mixture boils without change in composition at a definite temperature of an equimolar mixture of A and B is distilled, Where ${\mathrm{P}}_{\mathrm{A}}^0 \mathrm{and} {\mathrm{P}}_{\mathrm{B}}^0$ , $({\mathrm{P}}_{\mathrm{B}}^0>{\mathrm{P}}_{\mathrm{A}}^0)$ are vapour pressures of A and B respectively ${\mathrm{X}}_{\mathrm{B}}=$ mole fraction of B in liquid phase

Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $\mathrm{T} \left(\mathrm{K}\right)$ are $50 \mathrm{mmHg}$ and $40 \mathrm{mmHg}$, respectively. What is the mole fraction of toluene in vapour phase when $117 \mathrm{g}$ of benzene is mixed with $46 \mathrm{g}$ of toluene?

(Molar mass of benzene and toluene are $78$ and $92 \mathrm{g} {\mathrm{mol}}^{-1}$, respectively)

The correct option for the value of vapour pressure of a solution at $45º\mathrm{C}$ with benzene to octane in molar ratio $3 : 2$ is:

[At $45°\mathrm{C}$ vapour pressure of benzene is $280 \mathrm{mm} \mathrm{Hg}$ and that of octane is $420 \mathrm{mm} \mathrm{Hg}.$ Assume Ideal gas]