Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same
temperature, if 1 mol of A is farther added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is ____ . (Nearest integer)

To begin, we apply Raoult's law to find the vapour pressure of pure B.

For an ideal solution, the total vapour pressure is

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

where $$x_A, x_B$$ are mole fractions and $$P_A^*, P_B^*$$ are the vapour pressures of the pure components.

In the first mixture, there are 3 mol of A and 1 mol of B, giving a total of 4 mol. Thus

$$x_A = 3/4$$ and $$x_B = 1/4$$.

Substituting into Raoult's law gives

$$ \frac{3}{4}P_A^* + \frac{1}{4}P_B^* = 500 \quad \cdots (1) $$

After adding 1 mol of A, the amounts become 4 mol of A and 1 mol of B (total 5 mol), and the vapour pressure rises to 520 mm Hg.

The new mole fractions are $$x_A = 4/5$$ and $$x_B = 1/5$$.

Substituting again yields

$$ \frac{4}{5}P_A^* + \frac{1}{5}P_B^* = 520 \quad \cdots (2) $$

We now solve these equations simultaneously. Rewriting equation (1) gives

$$3P_A^* + P_B^* = 2000 \quad \cdots (1')$$

and equation (2) becomes

$$4P_A^* + P_B^* = 2600 \quad \cdots (2')$$

Subtracting (1') from (2') yields

$$ P_A^* = 2600 - 2000 = 600 \text{ mm Hg} $$

Substituting this value back into (1') gives

$$ 3(600) + P_B^* = 2000 $$

$$ 1800 + P_B^* = 2000 $$

$$ P_B^* = 200 \text{ mm Hg} $$

The answer is 200 mm Hg.