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 Concept:
Raoult's Law states that the total vapour pressure of a mixture of volatile liquids is the sum of their partial pressures.

Step 2: Formula Application:
$P_{total} = P_A + P_B = P^{\circ}_A X_A + P^{\circ}_B X_B$
Partial pressure of B ($P_B$) = $P_{total} \times Y_B$ (where $Y_B$ is mole fraction in vapour phase) OR if 900 is $P^{\circ}_B$, then $P_B = P^{\circ}_B X_B$. Given $P_B$ is 900? No, let's look at the logic: $P_B$ cannot be higher than $P_{total}$. Therefore, the "900" must be $P^{\circ}_B$.

Step 3: Explanation:
Given $X_B = 0.4$, so $X_A = 1 - 0.4 = 0.6$.
$P_{total} = P^{\circ}_A X_A + P^{\circ}_B X_B$
$600 = P^{\circ}_A(0.6) + (900)(0.4)$
$600 = 0.6 P^{\circ}_A + 360$
$240 = 0.6 P^{\circ}_A \implies P^{\circ}_A = 240 / 0.6 = 400$ mm Hg.

Step 4: Final Answer:
The vapour pressure of liquid A is 400 mm Hg.