Question

Vapour pressure of pure 'A' is 70 mm of Hg at $25^{\circ}C$. It forms an ideal solution with 'B' in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm of Hg at $25^{\circ}C$, the vapour pressure of pure 'B' at $25^{\circ}C$ is

• A. 28 mm
• B. 56 mm
• C. 70 mm
• D. 140 mm

Hint:

Raoult's Law: $P_{\text{total}} = P_A + P_B$. Always remember $X_A + X_B = 1$.

Answer 1

The Correct Option is D

Step 1: Apply Raoult's Law
For an ideal solution: $P = p_{A}^{\circ}x_{A} + p_{B}^{\circ}x_{B}$.
Step 2: Identify Values
$P = 84$, $p_{A}^{\circ} = 70$, $x_{A} = 0.8$. Thus, $x_{B} = 1 - 0.8 = 0.2$.
Step 3: Calculation
$84 = 70 \times 0.8 + p_{B}^{\circ} \times 0.2$.
$84 = 56 + 0.2 p_{B}^{\circ} \Rightarrow 0.2 p_{B}^{\circ} = 28$.
$p_{B}^{\circ} = \frac{28}{0.2} = 140$ mm.
Final Answer: (d)