Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 °C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 12.5 g of biphenyl in 27.9 g of benzene?

mole fraction benzene = XC6H6 = moles benzene/total mols.

mols benzene = grams/molar mass
mols biphenyl = grams/molar mass

Then pC6H6 = XC6H6 * PoC6H6 = ?

To find the vapor pressure of the solution, we need to use Raoult's Law. Raoult's Law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution.

First, let's determine the number of moles of biphenyl and benzene in the solution.

1. Calculate the number of moles of biphenyl (C12H10):
- Determine the molar mass of biphenyl:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of biphenyl = (12.01 * 12) + (1.01 * 10) = 154.24 g/mol
- Calculate moles of biphenyl:
Moles of biphenyl = 12.5 g / 154.24 g/mol

2. Calculate the number of moles of benzene (C6H6):
- Determine the molar mass of benzene:
Molar mass of C6H6 = (12.01 * 6) + (1.01 * 6) = 78.11 g/mol
- Calculate moles of benzene:
Moles of benzene = 27.9 g / 78.11 g/mol

Next, we can calculate the mole fractions of the solute and solvent:

1. Mole fraction of biphenyl:
Mole fraction of biphenyl = Moles of biphenyl / (Moles of biphenyl + Moles of benzene)

2. Mole fraction of benzene:
Mole fraction of benzene = Moles of benzene / (Moles of biphenyl + Moles of benzene)

Now, we can apply Raoult's Law to find the vapor pressure of the solution:

Vapor pressure of the solution = Mole fraction of benzene * Vapor pressure of pure benzene

Substitute the values into the equation to find the vapor pressure of the solution.