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 the total vapour pressure is $600$ $mm$ $Hg$. Also,find the composition of the vapour phase.

(A) Given:
$P_{A}^{0} = 450$ $mm$ $Hg$
$P_{B}^{0} = 700$ $mm$ $Hg$
$P_{total} = 600$ $mm$ $Hg$
According to Raoult's law:
$P_{total} = P_{A}^{0} x_{A} + P_{B}^{0} x_{B}$
Since $x_{B} = 1 - x_{A}$,we have:
$P_{total} = P_{A}^{0} x_{A} + P_{B}^{0} (1 - x_{A})$
$600 = 450 x_{A} + 700 (1 - x_{A})$
$600 = 450 x_{A} + 700 - 700 x_{A}$
$250 x_{A} = 100$
$x_{A} = \frac{100}{250} = 0.4$
$x_{B} = 1 - 0.4 = 0.6$
Composition of liquid mixture: $x_{A} = 0.4, x_{B} = 0.6$
Partial pressures:
$P_{A} = P_{A}^{0} x_{A} = 450 \times 0.4 = 180$ $mm$ $Hg$
$P_{B} = P_{B}^{0} x_{B} = 700 \times 0.6 = 420$ $mm$ $Hg$
Composition in vapour phase ($y_{A}$ and $y_{B}$):
$y_{A} = \frac{P_{A}}{P_{total}} = \frac{180}{600} = 0.3$
$y_{B} = \frac{P_{B}}{P_{total}} = \frac{420}{600} = 0.7$