The vapour pressure of pure liquids $A$ and $B$ are $450 \ mm \ Hg$ and $700 \ mm \ Hg$ respectively at $350 \ K$. Find out the composition of the liquid mixture if total vapour pressure is $600 \ mm \ Hg$. Also,find the composition of the vapour phase.

(N/A) Given: $P_A^0 = 450 \ mm \ Hg$,$P_B^0 = 700 \ mm \ Hg$,$P_{total} = 600 \ mm \ Hg$.
Let the mole fraction of $A$ in the liquid phase be $x_A$. Then the mole fraction of $B$ is $x_B = 1 - x_A$.
According to Raoult's Law: $P_{total} = P_A^0 x_A + P_B^0 x_B$.
$600 = 450 x_A + 700(1 - x_A)$.
$600 = 450 x_A + 700 - 700 x_A$.
$250 x_A = 100$.
$x_A = 100 / 250 = 0.4$.
So,$x_B = 1 - 0.4 = 0.6$.
The partial pressures are: $P_A = P_A^0 x_A = 450 \times 0.4 = 180 \ mm \ Hg$ and $P_B = P_B^0 x_B = 700 \times 0.6 = 420 \ mm \ Hg$.
In the vapour phase,the mole fractions are $y_A = P_A / P_{total} = 180 / 600 = 0.3$ and $y_B = P_B / P_{total} = 420 / 600 = 0.7$.