Question

Calculate the freezing point and boiling point of a solution containing 12.6 g of naphthalene (C10H8) in 109.0 mL of benzene. Benzene has a density of 0.877 g/cm3.

1. Calculate the freezing point of a solution.

2. Calculate the boiling point of a solution.

Answer 1

volume of benzene = 109.0 mL

mass of benzene = (volume of benzene) * (density of benzene)

mass of benzene = (109.0 mL) * (0.877 g/mL)

mass of benzene = 95.6 g

mass of benzene = 95.6 g * (1 kg / 1000 g)

mass of benzene = 0.0956 kg

mass naphthalene = 12.6 g

moles naphthalene = (mass naphthalene) / (molar mass naphthalene)

moles naphthalene = (12.6 g) / (128.17 g/mol)

moles naphthalene = 0.0983 mol

molality naphthalene = (moles naphthalene) / (mass benzene in kg)

molality naphthalene = (0.0983 mol) / (0.0956 kg)

molality naphthalene = 1.03m

(1.) decrease in freezing point = (K_f benzene) * (molality of naphthalene)

decrease in freezing point = (5.12 ^oC/m) * (1.03 m)

decrease in freezing point = 5.27 ^oC

Freezing point of solution = (normal freezing point of pure benzene) - (decrease in freezing point)

Freezing point of solution = (5.49 ^oC) - (5.27 ^oC)

Freezing point of solution = 0.225 ^oC

(2.) increase in boiling point = (K_b benzene) * (molality of naphthalene)

increase in boiling point = (2.53 ^oC/m) * (1.03 m)

increase in boiling point = 2.60 ^oC

boiling point of solution = (normal boiling point of pure benzene) + (increase in boiling point)

boiling point of solution = (80.1 ^oC) + (2.60 ^oC)

boiling point of solution = 82.7 ^oC