How do I find the total vapor pressure of a solution of benzene (mole fraction 0.350, vapor pressure at 25 degrees = 12.7 kPa) in toluene at 25 degrees. (vapor pressure of toluene at 25 degrees = 3.79 kPa)

Wouldnt you just multipy each vapor pressure x mole fraction of the component, then for total, add them up?

See Raoult's law of partial pressures.
http://en.wikipedia.org/wiki/Raoult's_law

To find the total vapor pressure of a solution, you can use Raoult's law, which states that the partial pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure.

In this case, you want to find the total vapor pressure of the solution of benzene (with a mole fraction of 0.350) in toluene at 25 degrees Celsius. The vapor pressure of benzene at 25 degrees Celsius is given as 12.7 kPa, and the vapor pressure of toluene at 25 degrees Celsius is given as 3.79 kPa.

To calculate the total vapor pressure, first find the partial pressure of benzene and toluene separately, and then add them together.

Partial pressure of benzene = mole fraction of benzene * vapor pressure of benzene
Partial pressure of benzene = 0.350 * 12.7 kPa = 4.445 kPa

Partial pressure of toluene = mole fraction of toluene * vapor pressure of toluene
Partial pressure of toluene = (1 - 0.350) * 3.79 kPa = 2.4645 kPa

Total vapor pressure of the solution = Partial pressure of benzene + Partial pressure of toluene
Total vapor pressure of the solution = 4.445 kPa + 2.4645 kPa = 6.9095 kPa

Therefore, the total vapor pressure of the solution of benzene in toluene at 25 degrees Celsius is approximately 6.9095 kPa.