A solution is prepared by mixing 8.5 g of $C{H}_{2}C{l}_2$ and 11.95 g of $CHC{l}_3$ at 298 K are 415 and 200mmHg respectively, the mole fraction of $CHC{l}_3$ in vapour form is :$(Molar mass of CI = 35.5g mo{l}^{-1}$)

Given data:
It is given to identify the mole fraction of $CHC{l}_3$ in vapour form.
Explanation:
$$\begin{gathered} P_0 =Vapor pressure = 200 mm of Hg \\ Mass of solute \left(C{H}_{2}C{l}_2\right)=8.5 g \\ Mass of solvent \left(CHC{l}_3\right)=11.95 g \\ The mole fraction in solution phase is computed as follows: \\ \begin{array}{l}{\mathrm{X}}_{{\mathrm{CH}}_2{\mathrm{Cl}}_2}=\frac{(8.5/85)}{(8.5/85)+(11.95/119.5)}=0.5\\ {\mathrm{X}}_{{\mathrm{CHCl}}_3}=\frac{(11.95/119.5)}{(8.5/85)+(11.95/119.5)}=0.5\end{array} \\ The formula for mole fraction in vapor phase is: \\ {\mathrm{Y}}_{{\mathrm{CHCl}}_3}=\frac{\mathrm{P}°·{\mathrm{X}}_{{\mathrm{CHCl}}_3}}{\mathrm{P}°·{\mathrm{X}}_{{\mathrm{CHCl}}_3}+\mathrm{P}°·{\mathrm{X}}_{{\mathrm{CH}}_2\mathrm{Cl}2}}=\frac{200\times 0.5}{200\times 0.5+415\times 0.5}=0.325 \end{gathered}$$

Hence, the correct option is option (D) 0.325.