The vapour pressures of benzene and toluene at `20^(@)C` are 75mmHg and 22mmHg respectively. `23.4g` of benzene and `64.4` g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase if the vapours are in equilibrium with the liquid mixture at the same temperature.

Answer

0.59

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.`