Question

Equal moles of benzene and toluene are mixed, the vapour pressure of benzene and toluene in pure state are 700 and 600 mm hg respectively. The mole fraction of benzene in vapour state is

Answer 1

This question is on Rault's law.

The total vapour pressure of a gaseous mixture is equal to the sum of vapour pressure of the individual gases.

Again, to get the partial vapour pressure all you need to do is dividing the mole of that given gas with the total number of moles of the gaseous mixture.

So when equal moles of benzene and toluene are mixed, the vapour pressure of benzene and toluene in pure state are 700 and 600 mm hg respectively.

The mole fraction of benzene is given by;

total number of moles =(700 + 600) = 1,300 mmHg

so the mole fraction is 700/ 1300

                                 =  0.5