The liquids X and Y from ideal solution having vapour pressures 200 and 100 mm Hg respectively. Calculate the mole fraction of component X in vapour phase in equilibrium with an equimolar solution of the two .

Answer

0.33

Explanation

In the liquid phase, `n_(B)=23.4//78=0.3, n_(T)=64.4//92=0.7." Hence, "x_(B)=0.3, x_(T)=0.7`
`p_(B)=0.3xx75=22.5 mm, p_(T)=0.7xx22=15.4 mm , "Tota vapour pressure "=22.5+15.4=37.9 mm"`
Mole fraction of benzene in the vapour phase `=22.5//37.9=0.59.`