Question

What is the mole fraction of benzene in the solution that has a vapor pressure of 38 torr at 20 ∘C? Express your answer using two significant figures.

Answer 1

The question is incomplete, here is the complete question:

At 20°C the vapor pressure of benzene $(C_6H_6)$ is 75 torr, and that of toluene $(C_7H_8)$ is 22 torr. Assume that benzene and toluene form an ideal solution.

What is the mole fraction of benzene in the solution that has a vapor pressure of 38 torr at 20°C? Express your answer using two significant figures.

Answer: The mole fraction of benzene is 0.302

Explanation:

Let the mole fraction of benzene be 'x' and that of toluene is '1-x'

To calculate the total pressure of the mixture of the gases, we use the equation given by Raoult's law, which is:

$p_T=\sum_{i=1}^n(\chi_{i}\times p_i)$

We are given:

Vapor pressure of benzene = 75 torr

Vapor pressure of toluene = 22 torr

Vapor pressure of solution = 38 torr

Putting values in above equation, we get:

$38=[(75\times x)+(22\times (1-x))]\\\\x=0.30$

Hence, the mole fraction of benzene is 0.30