Vapour pressure of chloroform `(CHCl_(3))` and dichloromethane `(CH_(2)Cl_(2))` at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) calculate the vapour pressure of the solution prepared by mixing 25.5 g of `CHCl_(3)` and 40g of `CH_(2)Cl_(2)` at 298 K and , (ii) mole fractions of each components in vapour phase.

Answer

(i) 348.1 mm Hg (ii) `x_(CHCl_(3))=0.18; x_(CH_(2)Cl_(2))=0.82`

Explanation

Molar mass of `CHCl_(3)=12+1+3xx35.5 =119.5` , Molar mass of `CH_(2)Cl_(2)=12+2xx1+2xx35.5=85`
Number of moles of `CHCl_(3)` in 25.5 g of `CHCl_(3)=(25.5)/(119.5)=0.213`
Number of moles of `CH_(2)Cl_(2)` in 40 g of `CH_(2)Cl_(2)=(40)/(85)=0.471`
Total number of moles in solution `=0.471 +0.213 = 0.684`
Mole fraction of `CHCl_(3)= (0.213)/(0.684)=0.311`
Mole fraction of `CH_(2)Cl_(2) =(0.471)/(0.684)=0.689`
Vapour pressure of `CHCl_(3)` in solution `= 200 xx 0.311 `
= 62.2 mm
Vapour pressure of `CH_(2)Cl_(3)` in solution `= 415 xx 0.689 = 285.9`,
Total vapour pressure `= 62.2+285.9=348.1` mm
Mole fraction of chloroform in vapour phase `=(62.2)/(348.1)=0.18`
Mole fraction of `CH_(2)Cl_(2)` in vapour phase `=(285.9)/(348.1)=0.82`