Benzene has a vapor pressure of 100.0 Torr at 26°C. A nonvolatile, nonelectrolyte compound was added to 0.309 mol benzene at 26°C and the vapor pressure of the benzene in the solution decreased to 60.0 Torr. What amount (in moles) of solute molecules were added to the benzene?

Isn't delta P = XsolutePosolvent

That is what I tried using...I solved for m for when P= 100 Torr, then used that to set up an equation when P=60 Torr to find the difference, but my answer is consistently incorrect. I am setting up the algebra incorrectly but am not sure what I am missing

This is not a molality problem. I think I've taken care of this problem further up the board.

To find the amount of solute molecules added to the benzene, we can use Raoult's law, which states that the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent in the solution.

The formula for Raoult's law is:

P_total = X_solvent * P_solvent

Where:
P_total is the total vapor pressure of the solution,
X_solvent is the mole fraction of the solvent in the solution,
P_solvent is the vapor pressure of the pure solvent.

First, let's calculate the mole fraction of the solvent in the solution:

Mole fraction of solvent (X_solvent) = Moles of solvent / Total moles of solution

Since we know the moles of benzene, we can calculate the total moles of the solution by adding the moles of the benzene to the moles of the solute.

Total moles of the solution = Moles of benzene + Moles of solute

Now, we can rearrange Raoult's law to solve for the moles of the solute:

Moles of solute = (P_solvent - P_total) * Total moles of solution / P_total

Using the given information, let's substitute the values into the equation:

P_total = 60.0 Torr (vapor pressure of the benzene in the solution)
P_solvent = 100.0 Torr (vapor pressure of benzene)
Total moles of solution = 0.309 mol (moles of benzene)

Moles of solute = (100.0 Torr - 60.0 Torr) * 0.309 mol / 60.0 Torr

Now we can calculate the moles of solute.