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 15.2 g of biphenyl in 29.5 g of benzene?
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 present.
First, we need to calculate the number of moles of each compound in the mixture:
Molar mass of biphenyl (C12H10) = 154.21 g/mol
Molar mass of benzene (C6H6) = 78.11 g/mol
Number of moles of biphenyl = mass / molar mass = 15.2 g / 154.21 g/mol = 0.0986 mol
Number of moles of benzene = mass / molar mass = 29.5 g / 78.11 g/mol = 0.3779 mol
Next, calculate the mole fraction of benzene (solvent) in the solution:
Mole fraction of benzene = moles of benzene / (moles of biphenyl + moles of benzene) = 0.3779 mol / (0.0986 mol + 0.3779 mol) = 0.7935
The mole fraction of biphenyl can be calculated by subtracting the mole fraction of benzene from 1:
Mole fraction of biphenyl = 1 - 0.7935 = 0.2065
Now, we can use Raoult's Law to calculate the vapor pressure of the solution:
Vapor pressure of the solution = mole fraction of benzene × vapor pressure of benzene
= 0.7935 × 100.84 torr
= 79.95 torr
Therefore, the vapor pressure of the solution is approximately 79.95 torr.