Question

The vapour pressure of a pure liquid A is 80 mmHg at 300K.It forms an ideal solution with liquid B.When the mole fraction of B is 0.4,the total pressure was found to be 88 mmHg.The vapour pressure of liquid A would be

Answer 1

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Answer 2

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The vapor pressure of a pure liquid A is 80 mmHg at 300K.It forms an ideal solution with liquid B. When the mole fraction of B is 0.4,the total pressure was found to be 88 mmHg. The vapor pressure of liquid B would be?

Answer:

The vapor pressure of liquid B would be $100mmHg$.

Explanation:

Given: Vapor pressure of pure liquid A, $P^{o}_{A}=80mmHg$

Total vapor pressure of solution , $P=88mmHg$

Consider $P^{o}_{B}$  is the vapor pressure of pure liquid B

Mole fraction of B, $x_{B}=0.4$

We know $x_{A}+x_{B}=1$

    $\\x_{A}=1=-0.4\\x_{A}=0.6$

According to Raoult's law:

$P=x_{A}P^{o}_{A}+x_{B}P^{o}_{B}$

$88=(0.6)(80)+(0.4)P^{o}_{B}$

$88=48+(0.4)P^{o}_{B}$

$P^{o}_{B}=100mmHg$