A solution is prepared by mixing 8.5 g ofCH₂CI₂ and 11.95 g of CHCl₃. If vapour pressure of ${\mathrm{CH}}_2{\mathrm{Cl}}_2\text{ and }{\mathrm{CHCl}}_3$ at 298K are 350 and 250 mmHg respectively, the mole fraction of CHCI₃ in vapour form is : (Molar mass of Cl = 35.5 g mol^-l)

Molar mass of CHCl₃ = 119.5 g/mol.
Molar mass of ${\mathrm{CH}}_2{\mathrm{Cl}}_2=85\mathrm{g}/\mathrm{mol}$
$\text{ Moles of }{\mathrm{CHCl}}_3=\frac{11.95}{119.5}=0.1\mathrm{mol}\text{. }$
$\text{ Moles of }{\mathrm{CH}}_2{\mathrm{Cl}}_2=\frac{8.5}{85}=0.1\mathrm{mol}\text{. }$
Mole fraction of CHCl₃ = $\frac{0.1}{0.2}=0.5\mathrm{mol}$
Mole fraction of CH₂Cl₂ = $\frac{0.1}{0.2}=0.5\mathrm{mol}$
(Given - Vapour pressure of ${\mathrm{CHCl}}_3=250\mathrm{mmHg}=0.329\text{ atm. }$
Vapour pressure of ${\mathrm{CH}}_2{\mathrm{Cl}}_2=350\mathrm{mmHg}=0.460\text{ atm.) }$(1 atm = 760mm Hg) 
$\therefore {\mathrm{P}}_{\text{(above solution) }}$
= Mole fraction of CHCl₃ x (vaporn pressure of CHCl₃) + Mole fraction of CH₂CI₂ x (vapour pressure of CH₂CI₂)
= 0.5 x 0.329+ 0.5 x 0.460 = 0.3945 
Mole fraction of CHCI₃ in vapour form $=\frac{0.1645}{0.3945}=0.4169$