Question

Biphenyl, C_12H_10, is a nonvolatile, nonionizing solute that is soluble in benzene, C_6H_6. At 25 degree C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 11.9 g of biphenyl in 28 4 g of benzene?

Answer 1

Vapour pressure of the solution containing a non-volatile solute is lower than the vapour pressure of pure solvent.

According to Raoults's law, the lowering of vapour pressure of a solvent is given by:

p* - p / p* = x₂

p* = vapour pressure of pure solvent

p = vapour pressure of solution

x₂ = mole fraction of solute

We have to calculate p.

x₂ = n₂ / n₁ + n₂

n = number of moles (1 stands for solvent and 2 for solute)

n = mass given / molecular mass

n₁ = 28.4 / 78.11 = 0.363

n₂ = 11.9 / 154.21 = 0.077

x₂ = 0.077 / 0.363 + 0.077 = 0.21

Given: p* = 100.84 torr

p* - p   = x₂ p*

100.84 - p = (0.21)(100.84)

100.84 - p = 21.176

100.84 - 21.176 = p

p = 79.66 torr

Vapour pressure of solution made from dissolving biphenyl in benzene is 79.66 torr