If 0.380 mol of a nonvolatile nonelectrolyte are dissolved in 3.00 mol of water, what is the vapor pressure P(H2O) of the resulting solution? The vapor pressure of pure water is 23.8 torr at 25 C.

Question

If 0.380 mol of a nonvolatile nonelectrolyte are dissolved in 3.00 mol of water, what is the vapor pressure P(H2O) of the resulting solution? The vapor pressure of pure water is 23.8 torr at 25 C.
I got this answer correct: 21.1 torr

A solution is composed of 1.50mol cyclohexane (Pey=97.6torr) and 2.20 mol acetone (P=229.5torr). What is the total vapor pressure (Ptotal) above this solution? 
I got the correct answer: 176 torr

However, I am confused about part B. Can anyone help explain how to do this problem?

As you saw in Part B, the vapor above the cyclohexane-acetone solution is composed of both cyclohexane vapor and acetone vapor. What mole fraction of the vapor above the solution (XeyVapor)is cyclohexane? 
i got the mole fraction of cyclohexane in the solution(.405) but that isn't what it wants and I don't know exactly what the question is asking for.

bobpursley · Answer

In the vapor phase,mole fraction will be in the same proportion as the pressure fractions.

Pey=97.6*1.50/3.7=40 about
Pac=229.5*2.2/3.7=136 about

Mole fraction ey= about 40/176=
Mole fraction vac=about 136/176

Sar · Answer

I see where i went wrong. I was trying to take the mole fraction (.405) and divide it into the 176. Thank you.