_{We have given two substances} $A and B$ _{with the pressure of their pure substances is in the ratio of}  $1: 2$

_i.e., $\frac{P_A^0}{P_B^0}=\frac{1}{2}$

_{Also, the mole ratio in solution is given}

_i.e., $\frac{X_A}{X_B}=\frac{1}{2}$

_{We know that Raoult's law gives us the relation between vapour pressure and concentration of the solution. It states that partial vapour pressure of each component is directly proportional to its mole fraction present in solution.}

_Therefore,

$P_A=P_A^0{X}_A$  _and    $P_B=P_B^0{X}_B$

_Where $P_A$  _and  $P_B$ _{is the partial vapour pressure of component}  $\mathrm{A}$  _and  $\mathrm{B}$_.

$P_A^0$  _and  $P_B^0$ _{is the vapour pressure in the pure state.}

$X_A$  _and  $X_B$ _{is its mole fraction.}

_{We know that   Dalton's law of partial pressure states that when a mixture of two or more non-reacting gases is enclosed in a container then the total pressure exerted by the gaseous mixture is equal to the sum of partial pressures of the individual gases.}

_{Now, from Dalton's law of partial pressure}

$P_{\text{total }}=P_A+P_B$

_{Substituting the values of} $P_A$ _and $P_B$_{, we get}

$P_{\text{total }}=P_A^0{X}_A+P_B^0{X}_B$

_{Also from Dalton's law, we know that the partial pressure of the gas in a mixture is the pressure which gas would exert when present alone in the same vessel at the same temperature as that in the mixture.}

_Therefore   $P_A=Y_A{P}_{\text{total }}$

$Y_A=\frac{P_A}{P_{\text{total }}}$

_{Substituting the values of}  $P_{\text{total }}$ _{and taking the reciprocate, we get}

$$\begin{gathered} \frac{1}{Y_A}=\frac{P_A^0{X}_A+P_B^0{X}_B}{P_A^0{X}_A} \\ \Rightarrow \frac{1}{Y_A}=1+\frac{P_B^0{X}_B}{P_A^0{X}_A} \end{gathered}$$

_{Now applying the values of given ratio, we get}

$$\begin{gathered} \Rightarrow \frac{1}{Y_A}=1+2\times 2=5 \\ \Rightarrow Y_A=\frac{1}{5}=0.2 \end{gathered}$$

_{The mole fraction of} $A$ _{in vapour pressure is} $0.2$.

_{Hence, option(d) is correct.}