[as0422]

Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene (C2HCl3) and trichloromethane (CHCl3) is established at 25 °C. The mole fraction of CHCl3 in the vapour phase is 0.73. What is the mass fraction of C2HCl3 in the liquid phase? Round your answer to two significant figures.

The vapour pressures of trichloroethene and trichloromethane at 25 °C are:

Pvap,C2HCl3 = 73.0 mmHg

Pvap,CHCl3 = 199.1 mm Hg

So, what I did was I found mole fraction of C2HCl3 and then used the two mole fractions along with the vapour pressures to find the total pressure of the solution.

P= (o.73)(199.1)+(0.27)(73) = 165.053mmHg

Then, from Raoult's Law I know that the mole fraction in liquid phase is equal to Mole fraction in vapour phase, multiplied by vapour pressure, divided by total pressure. From that, I found the mole fraction of both things in liquid phase. I use the mole fraction to find mass of both, and then did mass of C2HCl3 divided by the total mass that I calculated. I got an answer of

15.68/120=0.13 but it says that it's wrong. I'm not sure where I messed up?

[mjc123]

Your equation for P is wrong.
P = x_AP°_A + x_BP°_B
where x_A and x_B are the mole fractions in the liquid.
Can you use this equation to find an expression for x_A(vapour) in terms of x_A(liquid)?