Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionRaoult's LawMedium2 minQB

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Passage / Comprehension

The vapour pressure of two pure liquids, A and B that form an ideal solution are 300 and 800 torr respectively, at temperature T. A mixture of the vapour of A and B for which the mole fraction of A is 0.25 is slowly compressed at temperature T.

The mole-fraction of A in the last traces of vapours, is :

Options:

Answer:

Solution:

When the last traces of vapor remain, the mixture has completely condensed into liquid, making the liquid mole fractions $X_{A} = 0.25$ and $X_{B} = 0.75$ . The total pressure of the system at this point is calculated as $P = X_{A} P_{A}^{\circ} + X_{B} P_{B}^{\circ} = (0.25 \times 300) + (0.75 \times 800) = 675 t orr$ .

The mole fraction of A in the final traces of vapor ( $Y_{A}$ ) is determined by Dalton's Law: $Y_{A} = \frac{X _{A} P _{A}^{\circ}}{P}$ . Substituting the values yields $Y_{A} = \frac{0.25 \times 300}{675} = \frac{75}{675} = \frac{1}{9} \approx 0.11$ , which confirms Option (C).

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

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