Question

At ${90}^o$C, the vapour pressure of toluene is $400$ Torr and that of o-xylene is $150$ Torr. What is the composition of the liquid mixture that boils at ${90}^o$C when the pressure is $0.50$ atm? What is the composition of the vapour produced?

Answer 1

Given that

vapour pressure of toluene, $p_{toluene}=400$ Torr

vapour pressure of o-xylene, $p_{o-xylene}=150$ Torr at ${90}^o$C

let mixute has mole fraction of toluene as ${\chi }_t$ and that of o-xylene is ${\chi }_o$

pressure of the mixture$P_T=0.5atm=0.5\times 760=380torr$

using Raolt's law

$P_T={\chi }_t.p_{toluene}+{\chi }_o.p_{o-xylene}$

since ${\chi }_t=1-{\chi }_o$

$380={\chi }_t.400+(1-{\chi }_t)150$

${\chi }_t=\frac{380-150}{250}=0.92$

and ${\chi }_o=1-0.92=0.08$

molefraction of toluene is 0.92 and that of o-xylene is 0.08