Liquids A and B form ideal solution for all compositions of A and B at 25 $$^\circ C$$. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 bar, respectively. What is the vapor pressure of pure liquid B in bar?

For an ideal binary solution obeying Raoult’s law, the total vapour pressure $$P_{\text{tot}}$$ is

$$P_{\text{tot}} = x_A P_A^{\,0} + x_B P_B^{\,0}$$
where
$$x_A, x_B$$ are the mole fractions in the liquid (with $$x_A + x_B = 1$$) and
$$P_A^{\,0}, P_B^{\,0}$$ are the vapour pressures of the pure liquids A and B.

Data for the two given solutions

Solution 1: $$x_A = 0.25 \Rightarrow x_B = 0.75,\; P_{\text{tot}} = 0.30 \text{ bar}$$
Solution 2: $$x_A = 0.50 \Rightarrow x_B = 0.50,\; P_{\text{tot}} = 0.40 \text{ bar}$$

Write Raoult’s-law equations for the two solutions:

$$0.25\,P_A^{\,0} + 0.75\,P_B^{\,0} = 0.30 \quad -(1)$$
$$0.50\,P_A^{\,0} + 0.50\,P_B^{\,0} = 0.40 \quad -(2)$$

Multiply equation $$(1)$$ by 2 to eliminate $$P_A^{\,0}$$:

$$0.50\,P_A^{\,0} + 1.50\,P_B^{\,0} = 0.60 \quad -(1a)$$

Subtract $$(2)$$ from $$(1a)$$:

$$\bigl(0.50\,P_A^{\,0} - 0.50\,P_A^{\,0}\bigr) + \bigl(1.50\,P_B^{\,0} - 0.50\,P_B^{\,0}\bigr) = 0.60 - 0.40$$
$$1.00\,P_B^{\,0} = 0.20$$

Therefore, the vapour pressure of pure liquid B is

$$P_B^{\,0} = 0.20 \text{ bar}$$

Answer (numerical): 0.2 bar