A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution,

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Solution

Q.
A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are: [2025]

1 1400 mm Hg, A

2 1400 mm Hg, B

3 600 mm Hg, B

4 600 mm Hg, A

Ans.
(4)

Mole fraction of $A (X_{A}) = \frac{Moles of A}{Total moles} = \frac{1}{4}$

Mole fraction of $B (X_{B}) = \frac{Moles of B}{Total moles} = \frac{3}{4}$

By Raoult’s law, total pressure P is related to vapour pressure of pure A $(P_{A}^{°})$ and vapour pressure of pure B $(P_{B}^{°})$ as:

$P = P_{A}^{°} X_{A} + P_{B}^{°} X_{B}$

$500 = 200 \times \frac{1}{4} + P_{B}^{°} \times \frac{3}{4}$

$P_{B}^{°} = 600 mmHg$

As $P_{A}^{°} (200 mmHg) < P_{B}^{°} (600 mmHg)$ , A is the least volatile component.

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