Question

Calculate vapour pressure of volatile liquid A at given temperature if mole fraction and vapour pressure of volatile liquid B are 0.4 and 900 mm Hg respectively $[P_{\text{total}}=600\text{ mmHg}]$.

• A. 450 mm Hg
• B. 560 mm Hg
• C. 500 mm Hg
• D. 400 mm Hg

Hint:

Be careful reading the phrasing! When problems say "vapour pressure of volatile liquid B", they almost always mean the vapour pressure of the PURE liquid ($P^\circ_B$), not the partial pressure ($P_B$).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an ideal binary liquid mixture containing components A and B. We know the total vapour pressure, the mole fraction of B in the liquid phase, and the pure vapour pressure of B. We need to calculate the pure vapour pressure of component A ($P^\circ_A$).

Step 2: Detailed Explanation:
According to Raoult's Law for a mixture of volatile liquids, the total vapour pressure over the solution is the sum of the partial vapour pressures of its components:
$P_{\text{total}} = P_A + P_B$
$P_{\text{total}} = (P^\circ_A \times X_A) + (P^\circ_B \times X_B)$
We are given the following values:
Total pressure ($P_{\text{total}}$) = $600 \text{ mmHg}$
Mole fraction of liquid B ($X_B$) = $0.4$
Vapour pressure of pure liquid B ($P^\circ_B$) = $900 \text{ mmHg}$
Because this is a binary mixture containing only A and B, the sum of their liquid mole fractions must exactly equal 1:
$X_A + X_B = 1$
$X_A = 1 - 0.4 = 0.6$
Now, substitute all known values into Raoult's equation to solve for $P^\circ_A$:
$600 = (P^\circ_A \times 0.6) + (900 \times 0.4)$
$600 = 0.6 P^\circ_A + 360$
Rearrange the equation:
$600 - 360 = 0.6 P^\circ_A$
$240 = 0.6 P^\circ_A$
Solve for $P^\circ_A$:
$P^\circ_A = \frac{240}{0.6}$
$P^\circ_A = \frac{2400}{6} = 400 \text{ mmHg}$.

Step 3: Final Answer:
The vapour pressure of volatile liquid A is 400 mm Hg, matching option (d).