A solution is prepared by mixing 8.5 g of C H 2 C l 2 and 11.95 g of C H C l 3 . If vapour pressure of C H 2 C l 2 and C H C l 3 at 298 K are 415 and 200 mm Hg respectively, the mole fraction of C H C l 3 in vapour form is: M o l a r m a s s o f C l = 35.5 g m o l - 1

Question

A solution is prepared by mixing 8.5 g of C H 2 C l 2 and 11.95 g of C H C l 3 . If vapour pressure of C H 2 C l 2 and C H C l 3 at 298 K are 415 and 200 mm Hg respectively, the mole fraction of C H C l 3 in vapour form is: M o l a r m a s s o f C l = 35.5 g m o l - 1, 0.162, 0.675, 0.325, 0.486

MARKS App · Accepted Answer

mole of CH 2 C l 2 in liquid phase = 8.5 85 = .1 mole of CHCl 3 in liquid phase = 11.95 119.5 = .1 mole fraction of CH 2 Cl 2 in liquid phase = .1 .2 = 1 2 mole fraction of CHCl 3 in liquid phase = .1 .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 = 307.5 P T ( X A ) V P = ( X A ) L P × Vapour pressure of CHCl 3 307.5 × ( X C H C l 3 ) V P = 200 × 1 2 X C H C l 3 = 100 307.5 = .325

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