Question

At 25⁰Cthe vapor in equilibrium with a solution containing carbon disulfide and acetonitrile has a total pressure of 263 torr and is 85.5 mole percent carbon disulfide. What is the mole fraction of carbon disulfide in the solution? At 25⁰Cthe vapor pressure of carbon disulfide is 375 torr. Assume that the solution and the vapor exhibit ideal behavior.

Short answer

In the solution of carbon disulfide and acetonitrile, the mole fraction of carbon disulfide will be 59.9 mole percent or 0.599 mole.

Step-by-step solution

Determine the pressure

According to the question, it is given that

The total pressure of solution = 263 torr

The vapor pressure of carbon disulfide = 375 torr

Mole fraction of carbon disulfide = 85.5 mole percent or 0.855 mole fraction

Therefore, the partial pressure of carbon disulfide will be determined by

$\begin{array}{rcl}& & {\text{P}}_{{\text{CS}}_2}{\text{=x}}_{{\text{CS}}_2}\times {\text{P}}_{\text{Total}}\\ & & {\text{P}}_{{\text{CS}}_2}\text{=263}\times \text{\hspace{0.33em}0.855=224.9torr}\end{array}$

Determine mole fraction

Thus, the Mole fraction of carbon disulfide in a solution will be:

${\text{X}}_{\text{acetone}}\text{=}\frac{{\text{P}}_{{\text{CS}}_2}}{{{\text{P}}^{°}}_{{\text{CS}}_2}}\text{=}\frac{224.9}{375}\text{=0.599}$

Therefore, the mole fraction of carbon disulfide will be 59.9 %.