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…

A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mm Hg respectively, the mole fraction of CHCl3 in vapour form is: Molar mass of Cl=35.5 g mol-1
  1. 0.162
  2. 0.675
  3. 0.325
  4. 0.486

Solution

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 CHC l 3 ×  vapour pressure CHC l 3

=415× 1 2 +200× 1 2 =307.5

P T ( X A ) VP = ( X A ) LP ×   Vapour pressure of CHCl 3

307.5×  ( X CHC l 3 ) VP =200× 1 2

X CHC l 3 = 100 307.5 =.325

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Asked in: JEE-TOPICTESTS-CHEMISTRY

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