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Raoult's Law

Two liquids A...

Question

Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 2 mole of A and 3 mole of B is 500 mm Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increses by 20 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg);

p_{A}^{\circ} = 400 m m H g

and

p_{B}^{\circ} = 600 m m H g

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p_{A}^{\circ} = 300 m m H g

and

p_{B}^{\circ} = 700 m m H g

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p_{A}^{\circ} = 320 m m H g

and

p_{B}^{\circ} = 620 m m H g

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p_{A}^{\circ} = 430 m m H g

and

p_{B}^{\circ} = 680 m m H g

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Solution

The correct option is C $p_{A}^{\circ} = 320 m m H g$ and $p_{B}^{\circ} = 620 m m H g$
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, $X_{A} = \frac{2}{5}$
Mole fraction of B, $X_{B} = \frac{3}{5}$

Total vapour pressure of solution,
$= X_{A} . p_{A}^{\circ} + X_{B} . p_{B}^{\circ} 500 = \frac{2}{5} p_{A}^{\circ} + \frac{3}{5} p_{B}^{\circ} o r 2500 = 2 p_{A}^{\circ} + 3 p_{B}^{\circ} . . . . e q (i)$

After addition of $1 m o l$ of B,
Total vapour pressure of solution
( $2 m o l$ of A + $4 m o l$ B),
$= \frac{2}{6} p_{A}^{\circ} + \frac{4}{6} p_{B}^{\circ} 520 = \frac{2}{6} p_{A}^{\circ} + \frac{4}{6} p_{B}^{\circ} o r 3120 = 2 p_{A}^{\circ} + 4 p_{B}^{\circ} . . . . e q (i i)$
Solving equations (i) and (ii), we get,
$p_{B}^{\circ} = 620 m m H g \Rightarrow$ vapour pressure of pure 'B'
$p_{A}^{\circ} = 320 m m H g \Rightarrow$ vapour pressure of pure 'A'

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Raoult's Law

Standard XII Chemistry

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