Question

Calculate the freezing point and boiling point of a solution containing 16.4 g of naphthalene (C10H8) in 115.0 mL of benzene. Benzene has a density of 0.877g/cm3.

What is the freezing point and the boiling point of the solution?

Answer 1

mass of solute = 16.4 g

molar mass of napthalene = 128.17 g /mol

moles of solute = 16.4 / 128.17

                           = 0.128

benzene mass = 115 x 0.877 = 100.855 g =0.100855 kg

molality = moles ./ mass of solvent

molality = 0.128 / 0.100855

              = 1.27 m

Kf of benzene = 5.12 oC/ m

freezing point of benzene = 5.5 oC

delta Tf = kf x m

To - Tf = Kf x m

5.5 - Tf = 5.12 x 1.27

Tf = - 0.998 oC

freezing point = - 0.998 oC

Tb - 80.1 = 2.65 x 1.27

boiling point Tb = 83.45 oC