Given: Vapour pressure of pure liquid A ($p_A^0$) = 450 mm Hg, Vapour pressure of pure liquid B ($p_B^0$) = 700 mm Hg, Total vapour pressure ($p_{total}$) = 600 mm Hg

Step 1: Calculate composition of liquid mixture

Let $x_A$ = mole fraction of A in liquid phase Let $x_B$ = mole fraction of B in liquid phase

Since $x_A + x_B = 1$, therefore $x_B = 1 - x_A$

According to Raoult's law for volatile liquids:

$$p_{total} = p_A^0 \cdot x_A + p_B^0 \cdot x_B$$

Substituting values:

$$600 = 450 \times x_A + 700 \times (1 - x_A)$$ $$600 = 450 x_A + 700 - 700 x_A$$ $$600 = 700 - 250 x_A$$ $$250 x_A = 100$$ $$x_A = 0.4$$

Therefore, mole fraction of A in liquid = 0.4 and mole fraction of B in liquid = 0.6

Step 2: Calculate composition of vapour phase

Partial pressure of A: $p_A = p_A^0 \times x_A = 450 \times 0.4 = 180$ mm Hg Partial pressure of B: $p_B = p_B^0 \times x_B = 700 \times 0.6 = 420$ mm Hg

Mole fraction of A in vapour phase = $p_A / p_{total} = 180/600 = 0.3$ Mole fraction of B in vapour phase = $p_B / p_{total} = 420/600 = 0.7$

Answer: Liquid mixture composition: A = 0.4, B = 0.6; Vapour phase composition: A = 0.3, B = 0.7