1. Use Raoult’s law for an ideal solution

For a binary ideal solution of liquids $A$ and $B$:

$P_{\text{total}} = x_A P_A^0 + x_B P_B^0$

where

• $x_A$ and $x_B$ are mole fractions in liquid phase,
• $P_A^0$ and $P_B^0$ are vapour pressures of pure liquids $A$ and $B$.

Also,

$x_B = 1 - x_A$

2. Form equations from the given data

Case 1:

When $x_A = 0.25$, then $x_B = 0.75$ and total vapour pressure is $0.3$ bar.

So,

$0.25 P_A^0 + 0.75 P_B^0 = 0.3$

Case 2:

When $x_A = 0.50$, then $x_B = 0.50$ and total vapour pressure is $0.4$ bar.

So,

$0.50 P_A^0 + 0.50 P_B^0 = 0.4$

3. Solve the simultaneous equations

From the second equation:

$0.5(P_A^0 + P_B^0) = 0.4$

$P_A^0 + P_B^0 = 0.8$

So,

$P_A^0 = 0.8 - P_B^0$

Substitute into the first equation:

$0.25(0.8 - P_B^0) + 0.75P_B^0 = 0.3$

$0.2 - 0.25P_B^0 + 0.75P_B^0 = 0.3$

$0.2 + 0.5P_B^0 = 0.3$

$0.5P_B^0 = 0.1$

$P_B^0 = 0.2 \text{ bar}$

4. Final answer

The vapour pressure of pure liquid $B$ is:

$\boxed{0.2 \text{ bar}}$