A solution is prepared by mixing 8.5 g of CH_2 Cl_2 and 11.95 g of CHCl_3. If vapour pressure of CH_2 Cl_2 and CHCl_3 at 298 K are 415 and 200 mmHg…

A solution is prepared by mixing $\displaystyle 8.5$ g of $\displaystyle \mathrm{CH}_2 \mathrm{Cl}_2$ and $\displaystyle 11.95$ g of $\displaystyle \mathrm{CHCl}_3$. If vapour pressure of $\displaystyle \mathrm{CH}_2 \mathrm{Cl}_2$ and $\displaystyle \mathrm{CHCl}_3$ at $\displaystyle 298$ K are $\displaystyle 415$ and $\displaystyle 200$ mmHg respectively, the mole fraction of $\displaystyle \mathrm{CHCl}_3$ in vapour form is : (Molar mass of $\displaystyle \mathrm{Cl}=35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )