Two liquids A and B form an ideal solution at temperature TK. At TK, the vapour pressures of pure A and B are 55 and 15 kN $$m^{-2}$$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?

We need to find the mole fraction of A in the liquid phase that is in equilibrium with a vapor containing mole fraction of A = 0.8.

$$P_A^0 = 55$$ kN/m², $$P_B^0 = 15$$ kN/m²

Mole fraction of A in vapor: $$y_A = 0.8$$

Using Raoult's law and Dalton's law:

Total pressure: $$P_{total} = x_A P_A^0 + (1-x_A) P_B^0 = x_A(55) + (1-x_A)(15) = 40x_A + 15$$

Mole fraction in vapor:

$$y_A = \frac{x_A P_A^0}{P_{total}} = \frac{55x_A}{40x_A + 15}$$

Setting $$y_A = 0.8$$:

$$0.8 = \frac{55x_A}{40x_A + 15}$$

$$0.8(40x_A + 15) = 55x_A$$

$$32x_A + 12 = 55x_A$$

$$23x_A = 12$$

$$x_A = \frac{12}{23} = 0.5217$$

Therefore, the mole fraction of A is Option 4: 0.5217.