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 19.6 g of biphenyl in 25.1 g of benzene?

19.6 g of biphenyl /154.21 g of biphenyl*mol-1= moles of biphenyl

25.1 g of benzene/78.11 g of benzene *mol-1= moles of benzene

Since at 25°C, the vapor pressure of pure benzene is 100.84 torr, then the pressure must be equal to the mole fraction of benzene to the total number of moles in the solution

100.84 torr*(moles of benzene/[moles of benzene + moles of biphenyl])= pressure of the solution

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

First, we need to calculate the mole fraction of the solvent (benzene).

1. Calculate the number of moles of biphenyl:
- Determine the molar mass of biphenyl (C12H10):
12(12.01 g/mol) + 10(1.01 g/mol) = 154.22 g/mol
- Convert the mass of biphenyl to moles:
19.6 g / 154.22 g/mol = 0.127 moles

2. Calculate the number of moles of benzene:
- Determine the molar mass of benzene (C6H6):
6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
- Convert the mass of benzene to moles:
25.1 g / 78.11 g/mol = 0.321 moles

3. Calculate the mole fraction of benzene:
Mole fraction of benzene = moles of benzene / total moles
Mole fraction of benzene = 0.321 moles / (0.127 moles + 0.321 moles) = 0.716

Now we can use Raoult's law to find the vapor pressure of the solution.

The vapor pressure of the solution = mole fraction of benzene × vapor pressure of pure benzene

Vapor pressure of the solution = 0.716 × 100.84 torr = 72.18 torr

Therefore, the vapor pressure of the solution made from dissolving 19.6 g of biphenyl in 25.1 g of benzene is 72.18 torr.

mols biphenyl = grams/molar mass.

mols benzene = grams/molar mass.
Xbenzene = nbenzene/total mols.

Psoln = Xbenzene*Pobenzene