Question

Calculate the freezing point and boiling boing of a solution containing 11.4 g of naphthalene in 109.0 mL benzene.

Answer 1

benzene density = 0.877 g / mL

volume of benzene = 109.0 mL

mass of benzene =density x volume = 0.877 x 109 = 95.6g

solute mass = 11.4 g

molality = m = (11.4 / 128 ) x 1000 / 95.6

= 0.93 m

standard data :

Kb of benzene = 2.65 oC / m

Kf of benzene = 5.12 oC /m

freezing point of benzene = 5.5 oC

boiling point of benzene = 80.1 oC

Tf = Kf x m

To-Tf = 5.12 x 0.93

5.5 - Tf = 4.76

Tf = 0.74 oC

freezing point = 0.74 oC

Tb = Kb x m

Tb-To = 2.65 x 0.93

Tb - 80.1 = 2.46

Tb = 82.56 oC

boiling point point = 82.56 oC