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 19.9 g of biphenyl in 34.3 g of benzene?
mols biphenyl = grams/molar mass.
mols benzene = grams/molar mass
Xbenzene = mols benzene/total mols.
pbenzene soln = Xbenzene*Ppure benzene solvent.
To find the vapor pressure of the solution, we can use Raoult's Law, which states that the vapor pressure of an ideal solution is equal to the product of the mole fraction of the solute and the vapor pressure of the pure solvent.
First, let's calculate the number of moles of biphenyl and benzene in the solution.
The molar mass of biphenyl (C12H10) is:
12 * 12.01 (carbon) + 10 * 1.01 (hydrogen) = 154.22 g/mol.
To find the moles of biphenyl:
moles of biphenyl = mass of biphenyl / molar mass of biphenyl
moles of biphenyl = 19.9 g / 154.22 g/mol
Next, we calculate the moles of benzene:
molar mass of benzene (C6H6) = 6 * 12.01 (carbon) + 6 * 1.01 (hydrogen) = 78.11 g/mol
moles of benzene = mass of benzene / molar mass of benzene
moles of benzene = 34.3 g / 78.11 g/mol
Now, we can calculate the mole fraction of benzene:
mole fraction of benzene = moles of benzene / (moles of biphenyl + moles of benzene)
Next, we need to find the vapor pressure of the pure benzene (P°benzene), which is given as 100.84 torr.
Finally, we can use Raoult's Law to find the vapor pressure of the solution (Psolution):
Psolution = mole fraction of benzene * P°benzene
So, by plugging in the known values, we can find the answer.