Total vapour pressure of mixture of 1 mol of volatile component A ( p A ∘ = 100 mm Hg ) and 3 mol of volatile component B ( p B ∘ = 60 mm Hg ) is 75 mm . For such case :

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ChemistryLiquid SolutionRaoult's LawMedium2 minQB

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Answer the questions (given below) which are based on the following diagram. Consider some facts about the above phase diagram : Vapour pressure diagram for real solutions of two liquids A and B that exhibit a positive deviation from Raoult's law. The vapour pressure of both A and B are greater than predicted by Raoult's law. The dashed lines represented the plots for ideal solutions.

Total vapour pressure of mixture of 1 mol of volatile component $A (p_{A}^{\circ} = 100 mm Hg)$ and 3 mol of volatile component $B (p_{B}^{\circ} = 60 mm Hg)$ is 75 mm . For such case :

Options:

Answer:

Solution:

$X_{A} P_{A}^{0} + X_{B} P_{B}^{0} = \frac{1}{4} \times 100 + \frac{3}{4} \times 60 = 70 mm$ of Hg

Now $P_{A} + P_{B} > X_{A} P_{A}^{0} + X_{B} P_{B}^{0}$

i.e., positive deviation from Raoult's law.

for positive deviation, B.P. < expected value

$F_{A \dots B} < F_{A \dots A}$

$F_{A \dots B} < F_{B \dots B}$

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

⏱ 2mℹ️ Source: QB

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