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 11.1 g of biphenyl in 29.1 g of benzene?
mols biphenyl = grams/molar mass.
mols benzene = grams/molar mass.
Xbenzene = mols benzene/total mols
Psoln = Xbenzene*Pobenzene
To find the vapor pressure of the solution, we can use Raoult's Law.
According to Raoult's Law, the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent present in the solution.
First, we need to calculate the mole fraction of benzene in the solution.
Moles of benzene = mass of benzene / molar mass of benzene
Molar mass of benzene (C6H6) = (6 * atomic mass of carbon) + (6 * atomic mass of hydrogen)
= (6 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 78.11 g/mol
Moles of benzene = 29.1 g / 78.11 g/mol
Next, we calculate the mole fraction of biphenyl in the solution.
Moles of biphenyl = mass of biphenyl / molar mass of biphenyl
Molar mass of biphenyl (C12H10) = (12 * atomic mass of carbon) + (10 * atomic mass of hydrogen)
= (12 * 12.01 g/mol) + (10 * 1.01 g/mol)
= 154.22 g/mol
Moles of biphenyl = 11.1 g / 154.22 g/mol
Now, calculate the total moles in the solution:
Total moles = moles of benzene + moles of biphenyl
Finally, calculate the mole fraction of benzene (Xbenzene) and biphenyl (Xbiphenyl):
Xbenzene = moles of benzene / total moles
Xbiphenyl = moles of biphenyl / total moles
Now, we can use Raoult's Law to calculate the vapor pressure of the solution.
Vapor pressure of the solution = Xbenzene * vapor pressure of pure benzene
Substituting the values:
Vapor pressure of the solution = Xbenzene * 100.84 torr
By calculating the mole fractions, plugging in the values, and performing the multiplication, you can determine the vapor pressure of the solution made from dissolving 11.1 g of biphenyl in 29.1 g of benzene at 25 °C.