Finding the mole fraction

Question

At $25\ \mathrm{^\circ C}$ the vapor in equilibrium with a solution containing carbon disulfide and acetonitrile has a total pressure of $263\ \mathrm{torr}$ and is $85.5\ \mathrm{mol}\text{-}\%$ carbon disulfide. What is the mole fraction of carbon disulfide in the solution? At $25\ \mathrm{^\circ C}$ the vapor pressure of carbon disulfide is $375\ \mathrm{torr}$. Assume that the solution and the vapor exhibit ideal behavior.

I was gonna use Raoult's law but like how is the molar percentage different from the molar fraction, isnt it the molar fraction times 100? also I checked the solution manual and it had the molar fraction sign but with an L or V above it, and theres no such thing in the boook, can anyone help or explain what is that?

Answer 1

As you said, the molar percentage is just the molar fraction multiplied by 100. As to the L and V above the molar fractions, they refer to the molar fraction of the component in the liquid and vapor phase, respectively. They are needed since (in general) the molar fraction of a component in a mixture are different in the liquid and vapor phase.

So, from the data in your problem, you know that the molar fraction of carbon disulfide in the gas phase is 0.855, and let's call it $x_{CS_2}^V$. The total pressure is P=263 torr, and since the vapor exhibits ideal behavior, the partial pressure of CS2 is: $P_{CS_2} =x_{CS_2}^V P$. Using Raoult's law (since the liquid exhibits ideal behavior), this partial pressure must be equal to the molar fraction of CS2 in the liquid times the vapor pressure of pure CS2: $P_{CS_2} =x_{CS_2}^L P_{CS_2}^o$. And I am sure you can follow from here.