It is given that:

${\mathrm{p}}_{\mathrm{A}}^0$= 450 mm of Hg

${\mathrm{p}}_{\mathrm{B}}^0$= 700 mm of Hg

p_total = 600 mm of Hg

From Raoult’s law, we have:

${\mathrm{p}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}$

${\mathrm{p}}_{\mathrm{B}}={\mathrm{p}}_{\mathrm{B}}^0{\mathrm{x}}_{\mathrm{B}}={\mathrm{p}}_{\mathrm{B}}^0(1-{\mathrm{x}}_{\mathrm{A}})$

Therefore, total pressure,

${\mathrm{p}}_{\mathrm{total}}={\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}$
$\Rightarrow {\mathrm{p}}_{\mathrm{total}}={\mathrm{p}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}^0(1-{\mathrm{x}}_{\mathrm{A}})$
$\Rightarrow {\mathrm{p}}_{\mathrm{total}=}{\mathrm{p}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}^0-{\mathrm{p}}_{\mathrm{B}}^0{\mathrm{x}}_{\mathrm{A}}$
$\Rightarrow {\mathrm{p}}_{\mathrm{total}}=({\mathrm{p}}_{\mathrm{A}}^0-{\mathrm{p}}_{\mathrm{B}}^0){\mathrm{x}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}^0$
$\Rightarrow 600=(450-700){\mathrm{x}}_{\mathrm{A}}+700$
$\Rightarrow -100=-250{\mathrm{x}}_{\mathrm{A}}$
$\Rightarrow {\mathrm{x}}_{\mathrm{A}}=0.4$

Therefore,

${\mathrm{x}}_{\mathrm{B}}=1-{\mathrm{x}}_{\mathrm{A}}$

= 1 − 0.4

= 0.6

Now, ${\mathrm{p}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}=450\times 0.4=180<mspace\; width="1em"></mspace>\mathrm{mm}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{Hg}$ 

${\mathrm{p}}_{\mathrm{B}}={\mathrm{p}}_{\mathrm{B}}^0{\mathrm{x}}_{\mathrm{B}}=700\times 0.6=420<mspace\; width="1em"></mspace>\mathrm{mm}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{Hg}$

Now, in the vapour phase:

Mole fraction of liquid A$=\frac{{\mathrm{p}}_{\mathrm{A}}}{{\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}}$

$=\frac{180}{180+420}$
$=\frac{180}{600}$

= 0.30

And, mole fraction of liquid B = 1 − 0.30 = 0.70