If two substance A and B have ${\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}:{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}=1$ : 2 and also x_A : x_B in solutins as 1 : 2, then mole fraction of A in vapours is

Given, two substances A and B 
We need to find the mole fraction of A in vapours is, from the given option.

$$\begin{gathered} \frac{{\mathrm{P}}_A°}{{\mathrm{P}}_{\mathrm{B}}°}=\frac{1}{2} \\ \text{ Let, }{\mathrm{P}}_{\mathrm{A}}°=1 \\ {\mathrm{P}}_{\mathrm{B}}^0=2 \\ Likewise, n_A=1 and n_B =2 \\ By Raoult\text{'}s law, \\ \begin{array}{l}{\mathrm{P}}_{\mathrm{A}}={\mathrm{X}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{A}}°\\ \therefore {\mathrm{P}}_{\mathrm{A}}=\frac{1}{3}\times 1=\frac{1}{3}\\ \& {\mathrm{P}}_{\mathrm{B}}={\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{B}}°\\ =\frac{2}{3}\times 2=\frac{4}{3}\\ {\mathrm{P}}_{\mathrm{A}}={\mathrm{X}}_{\mathrm{A}}\times {\mathrm{P}}_{\mathrm{T}}\\ \therefore \frac{1}{3}={\mathrm{X}}_{\mathrm{A}}\times \left(\frac{1}{3}+\frac{4}{3}\right)\\ \therefore {\mathrm{X}}_{\mathrm{A}}=\frac{1/3}{5/3}=\frac{1}{5}=0.2\end{array} \end{gathered}$$

Therefore the correct option is D 0.2