Question

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 12.3 g of biphenyl in 28.2 g of benzene?

Answer 1

As per Raoult's Law,

fractional change in vapour pressure = mole fraction of the non-volatile solute

Thus, (P⁰ - P)/P⁰ = X ; where P⁰ = vapor pressure of pure benzene, P = Pressure of the solution, X = mole fraction of the solute

Now, mole fraction of solute = moles of solute/(moles of solute+moles of benzene)

Molar mass of solute = 154 g/mole

molar mass of benzene = 78 g/mole

Thus, moles of solute = mass/molar mass = 12.3/154 = 0.08

moles of benzene = 28.2/78 = 0.36

Thus, mole fraction of solute = 0.08/(0.08+0.36) = 0.182

Thus, P = vapor pressure of solution = 100.84 - 0.182*100.84 = 82.505 torr