Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionRaoult's LawMedium2 minQB

ChemistryMediumsingle choice

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 pressure at which first drop of liquid will form, is :

Options:

Answer:

Solution:

Given vapor composition:

$y_{A} = 0.25$ , $y_{B} = 0.75$

At the first drop of liquid formation (dew point):

$P = \frac{1}{\frac{y _{A}}{P _{A}^{0}} + \frac{y _{B}}{P _{B}^{0}}}$

Substitute values:

$P = \frac{1}{\frac{0.25}{300} + \frac{0.75}{800}}$

$P = \frac{1}{0.000833 + 0.000937} = 564.7 t orr$

Therefore, the pressure at which the first drop of liquid forms is $564.7 t orr$ .

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

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...