At 298 K the vapour pressure of pure benzene `C_(6) H_(6) ` is 0.256 bar and vapour pressure of pure toluene, `C_(6)H_(8) ` is 0.925 bar. If the mole fraction of benzene in solution is 0 . 40 , find the total vapour pressure of solution. Also find the mole fraction of toluene in vapour phase

Answer

Total vapour pressure = 0.6574 bar . Molar fraction of toluene in vapour phase =0.844

Explanation

` x_(B) = 0 . 40`
`:. x_(T) = 1 - x_(B)`
`= 1 - 0 . 4 = 0 . 6 `
`p_(B)^(@) = 0 . 256 b a r , p_(T) ^(@)`
= . 925 bar
`:. p_(B) = p_(B) + p_(T)`
`= p_(B)^(@) x_(B) + p_(T)^(@) x_(T)`
`= 0 . 256 xx 0 . 40 + 0 . 925 xx 0 . 6`
`= 0 . 1024 + 0 .555`
= 0 .6574 bar
Mole fraction of toluene in Vpour phase
`y_(T) = (p_(T))/(p_("solution")) = (p_(T)^(@) x_(T))/( p_("solution "))`
`= (0 . 555)/( -0. 6574) = 0 . 844 `