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 mm Hg respectively, the mole fraction of CHCl 3 in vapour form is: M o l a r m a s s o f Cl = 35.5 g m o l - 1

Step-by-step solution

mole of CH 2 C l 2   in liquid phase = 8.5 85 = 0.1 mole of CHCl 3   in liquid phase = 11.95 119.5 = 0.1 mole fraction of CH 2 Cl 2   in liquid phase = 0.1 0.2 = 1 2 mole fraction of CHCl 3   in liquid phase = 0.1 0.2 = 1 2 P T = X C H 2 C l 2 ×     vapour pressure C H 2 C l 2 +   X C H C l 3 ×   vapour pressure C H C l 3 = 415 × 1 2 + 200 × 1 2 = 30