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.

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

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

Thank you.

To find the mole fraction of cyclohexane in the vapor above the solution (XcyclohexaneVapor), you need to use the vapor pressure of cyclohexane (Pey) and the total vapor pressure above the solution (Ptotal).

The mole fraction of cyclohexane in the vapor can be calculated using Raoult's law, which states that the partial pressure of a component in a solution is equal to the product of its mole fraction in the vapor phase and the vapor pressure of that component:

PcyclohexaneVapor = XcyclohexaneVapor * Pey

Since the total vapor pressure above the solution (Ptotal) is the sum of the partial pressures of cyclohexane and acetone:

Ptotal = PcyclohexaneVapor + PacetoneVapor

We can rearrange the equation to solve for XcyclohexaneVapor:

XcyclohexaneVapor = (PcyclohexaneVapor / Ptotal)

To calculate XcyclohexaneVapor, plug in the values for Pey (97.6 torr) and Ptotal (176 torr) into the equation:

XcyclohexaneVapor = (97.6 torr / 176 torr)

Calculating this gives us approximately 0.5545. Therefore, the mole fraction of cyclohexane in the vapor above the solution is 0.5545.

To find the mole fraction of cyclohexane in the vapor above the solution, you need to calculate the partial vapor pressure of cyclohexane (Pcyclohexane) and the partial vapor pressure of acetone (Pacetone), and then use the mole fractions to calculate the mole fraction of cyclohexane in the vapor.

First, calculate the partial vapor pressures using Raoult's law, which states that the vapor pressure of a component in a solution is directly proportional to its mole fraction in the solution.

Pcyclohexane = Xcyclohexane * P°cyclohexane
Pacetone = Xacetone * P°acetone

Where:
- Xcyclohexane and Xacetone are the mole fractions of cyclohexane and acetone in the solution.
- P°cyclohexane and P°acetone are the vapor pressures of pure cyclohexane and acetone, respectively.

Then, calculate the total vapor pressure above the solution (Ptotal) using Dalton's law of partial pressures, which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases.

Ptotal = Pcyclohexane + Pacetone

Finally, calculate the mole fraction of cyclohexane in the vapor (XcyclohexaneVapor) using the following equation:

XcyclohexaneVapor = Pcyclohexane / Ptotal

This will give you the mole fraction of cyclohexane in the vapor above the solution.

Make sure to substitute the correct values for Xcyclohexane, Xacetone, P°cyclohexane, P°acetone, Pcyclohexane, and Pacetone to obtain the accurate result.