A quantity of benzene, C6H6
and toluene, C6H5CH3
,is placed in a 1 L
evacuated vessel at 25 ∘C
At equilibrium, a small volume of liquid is visible at the bottom of the container. A sample of the vapour phase is analysed and found to contain 53 mol−%
benzene. What is the mole fraction of benzene in the liquid phase?
a. Vapour pressures at 25 ∘C
b. Pvap(benzene)=0.125 atm
c. Pvap(toluene)=0.037 atm
First of all, I am not even certain if this a Raoult's law problem. I just assumed so, since Henry's constant was not given in the problem.
First I'm getting the total liquid pressure using Raoult's law.
P(benzene)P(toluene)Ptotal=X(benzene)⋅Pvap(benzene)=X(toluene)⋅Pvap(toluene)=0.53×0.125=0.47×0.037=0.06625+0.01739=0.06625 atm=0.01739 atm=0.08364 atm
Then using Dalton's law, I am getting the mole fraction of Benzene in the liquid phase.
P(benzene)Y(benzene)=Y(benzene)⋅Ptotal=P(benzene)Ptotal=0.066250.08364=0.792.
However, my answer is incorrect according to the answer key that has 0.25
as the answer. I have a feeling my whole approach may be wrong since my answer is off by a lot. What am I doing wrong?
solutionsgas-laws
