A mixture contains 1 mole of volatile liquid A ( P A ∘ = 100 mm Hg ) and 3 moles of volatile liquid B ( P B ∘ = 80 mm Hg ) . If solution behaves ideally, the total vapour pressure of the distillat

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A mixture contains 1 mole of volatile liquid A ( P A ∘ ​ = 100 mm Hg ) and 3 moles of volatile liquid B ( P B ∘ ​ = 80 mm Hg ) . If solution behaves ideally, the total vapour pressure of the distillate is

TestHub · Accepted Answer

The total pressure of a mixture of 2 volatile liquids is given by:

$P_{T o t a l} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}$

$P_{T o t a l} = (100 \times \frac{1}{4}) + (80 \times \frac{3}{4}) = (25 + 60) = 85 mm H g$

The mole fraction of A in the vapor phase ( $X_{A}^{'}$ ) is given by:

$X_{A}^{'} = \frac{P _{A}^{\circ} X _{A}}{P _{T o t a l}} = \frac{100 \times \frac{1}{4}}{85} = \frac{25}{85} = 0.294$

Therefore, $X_{B}^{'} = 0.706$ .

These mole fractions will be present in the distillate. The total vapor pressure of the distillate would be:

$P_{T o t a l}^{'} = P_{A}^{\circ} X_{A}^{'} + P_{B}^{\circ} X_{B}^{'} = (100 \times 0.294) + (80 \times 0.706) = (29.4 + 56.48)$

$= 85.88 mm H g$

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