Question

Two liquids A and B are present in a solution: Vapour pressure of A is 70 mm Hg at 300 K, total vapour pressure of solution is 84 mm Hg. If mole fraction of B is 0.2, find vapour pressure of pure B at 300 K.

• A. 56 mm Hg
• B. 70 mm Hg
• C. 84 mm Hg
• D. 140 mm Hg

Answer 1

Answer: (D)

140 mm Hg

Answer 2

Given the total vapour pressure:

$$P_{\text{total}} = x_A P^*_A + x_B P^*_B$$

where:

• $P^*_A = 70$ mm Hg,

• $x_B = 0.2$ $\Rightarrow x_A = 1 - 0.2 = 0.8$,

• $P_{\text{total}} = 84$ mm Hg.

Substitute the known values:

$$84 = 0.8 \times 70 + 0.2 \times P^*_B$$ $$84 = 56 + 0.2\, P^*_B$$

Subtracting 56 from both sides:

$$28 = 0.2\, P^*_B$$

Now, solve for $P^*_B$:

$$P^*_B = \frac{28}{0.2} = 140 \text{ mm Hg}$$