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 B in the solution, whose normal boiling point is T, is :

Options:

Answer:

Solution:

Given:

$P_{A}^{0} = 300 t orr$ , $P_{B}^{0} = 800 t orr$

At the normal boiling point, the total vapor pressure equals the atmospheric pressure:

$P_{t o t a l} = 760 t orr$

For an ideal solution:

$P = X_{A} P_{A}^{0} + X_{B} P_{B}^{0}$

$760 = 300 X_{A} + 800 X_{B}$

Since $X_{A} + X_{B} = 1$ ,

$760 = 300 (1 - X_{B}) + 800 X_{B}$

$460 = 500 X_{B} \Rightarrow X_{B} = 0.92$

Therefore, the mole fraction of B in the solution is 0.92.

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

⏱ 2mℹ️ Source: QB

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