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…

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 298K 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 CH2Cl2 in liquid phase=8.585=0.1

mole of CHCl3 in liquid phase=11.95119.5=0.1

mole fraction of CH2Cl2 in liquid phase=0.10.2=12

mole fraction of CHCl3 in liquid phase=0.10.2=12

PT=XCH2Cl2×  vapour pressureCH2Cl2+ XCHCl3× vapour pressureCHCl3

=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

XCHCl3=100307.5=0.325

 

Asked in: JEE Main 2017 (09 Apr Online)

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