Given $: P^{0}_{A}=450 \text{ mm Hg},$ $P^{0}_{B}=700 \text{ mm Hg},$ $P_{\text{Total}}=600 \text{ mm Hg}, x_{A}=?$Applying Raoult's law, $P_{A}=x_{A} \times P^{0}_{A}$ $P_{B}=x_{B} \times P^{0}_{B} = (1-x_{A}) P^{0}_{B}$$P_{\text{Total}} = P_{A}+P_{B} = x_{A} \times P^{0}_{A} + (1-x_{A}) P^{0}_{B}$$= P^{0}_{B} + (P^{0}_{A} - P^{0}_{B}) x_{A}$Substituting the given values, we get$600=700+(450-700) x_{A}$ or, $250 x_{A}=100$ or $x_{A} = \frac{100}{250} = 0.40$Thus, composition of the liquid mixture will be$x_{A}($ mole fraction of $A)=0.40$$x_{B}($ mole fraction of $B)=1-0.40=0.60$Calculation of composition in the vapour phase$P_{A}=x_{A} \times P^{0}_{A}=0.40 \times 450 \text{ mm Hg}$ $=180 \text{ mm Hg}$$P_{B}=x_{B} \times P^{0}_{B}=0.60 \times 700 \text{ mm Hg}$ $=420 \text{ mm Hg}$Mole fraction of $A$ in the vapour phase $= \frac{P_{A}}{P_{A}+P_{B}}$$= \frac{180}{180+420}=0.30$Mole fraction of B in the vapour phase $= 1 - 0.30 = 0.70$