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Raoult's Law

Two liquids '...

Question

Two liquids 'A' and 'B' form an ideal solution. At $300 K$ , the vapour pressure of a solution containing $1 m o l$ of A and $3 m o l$ of B is $550 m m H g$ . At the same temperature if one more mole of B is added to this solution, the vapour pressure of the solution increases by $10 m m H g$ . Determine the vapour pressure of 'A' and 'B' in their pure states.

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Solution

Let the vapour pressure of pure 'A' be $p_{A}^{\circ}$ and the vapour pressure of pure 'B' be $p_{B}^{\circ}$
Initially,
Mole fraction of A, $X_{A} = \frac{1}{4}$
Mole fraction of B, $X_{B} = \frac{3}{4}$

Total vapour pressure of solution,
$= X_{A} . p_{A}^{\circ} + X_{B} . p_{B}^{\circ} 550 = \frac{1}{4} p_{A}^{\circ} + \frac{3}{4} p_{B}^{\circ} o r 2200 = p_{A}^{\circ} + 3 p_{B}^{\circ} . . . . e q (i)$

After addition of $1 m o l$ of B,
Total vapour pressure of solution
( $1 m o l$ of A + $4 m o l$ B),
$= \frac{1}{5} p_{A}^{\circ} + \frac{4}{5} p_{B}^{\circ} 560 = \frac{1}{5} p_{A}^{\circ} + \frac{4}{5} p_{B}^{\circ} o r 2800 = p_{A}^{\circ} + 4 p_{B}^{\circ} . . . . e q (i i)$
Solving equations (i) and (ii), we get,
$p_{B}^{\circ} = 600 m m H g \Rightarrow$ vapour pressure of pure 'B'
$p_{A}^{\circ} = 400 m m H g \Rightarrow$ vapour pressure of pure 'A'

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