Calculate vapour pressure of distillate of the solution at 2 5 ∘ C temperature.(in torr)

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ChemistryLiquid SolutionRaoult's LawMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

2 moles of a volatile liquid (A) and 3 moles of another volatile liquid (B) are present in a solution at 25°C temperature in a closed container if vapor pressure of pure A and pure B at 25°C are $P_{A}^{0} = 300 t orr$ and $P_{B}^{0} = 200 t orr$ , respectively (the above solution follows Raoult's law of partial vapor pressure).

Calculate vapour pressure of distillate of the solution at $2 5^{\circ} C$ temperature.(in torr)

Answer:

250.00

Solution:

$\frac{Y _{A}}{Y _{B}} = \frac{P _{A}^{0} X _{A}}{P _{B}^{0} X _{B}} = \frac{300 \times 2}{200 \times 3} = \frac{1}{1}$

$Y_{A} = 1/2 Y_{B} = 1/2$

These mole fractions are present in the distillate.

$P_{T} = P_{A}^{o} X_{A}^{'} + P_{B}^{o} X_{B}^{'} = 300 \times \frac{1}{2} + 200 \times \frac{1}{2} = 250 torr.$

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Raoult's Law

⏱ 2mℹ️ Source: QB

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