Vapour pressure of chloroform ( CHCl ₃ ) and dichloromethane ( CH ₂ Cl ₂ ) at 25^ C are 200 ~mm Hg and 415 ~mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 ~g of CHCl ₃ and 4 O g of CH ₂ Cl ₂ at the same temperature will be: [Molecular mass of CHCl ₃=119.5 ~g / mol and molecular mass of CH ₂ Cl ₂=85 ~g / mol ]

Step-by-step solution

Molar mass of CH ₂ Cl ₂=12 1+1 2+35.5 2=85 ~g ~mol ⁻¹ Molar mass of CHCl ₃=12 1+1 1+35.5 3=119.5 ~g ~mol ⁻¹ Moles of CH ₂ Cl ₂=40 ~g / 85 ~g ~mol ⁻¹=0.47 ~mol Moles of CHCl ₃=25.5 ~g / 119.5 ~g ~mol ⁻¹=0.213 ~mol Total number of moles =0.47+0.213=0.683 ~mol Mole fraction of component 2 =0.47 ~mol / 0.683 ~mol =0.688 Mole fraction of component 1 =1.00-0.688=0.312 We know that: aligned & P _ T = p ₁^0+ ( p ₂^0- p ₁^0 ) x ₂ & =200+(415-200) 0.688 & =200+147.9 & =347.9 ~mm Hg aligned