Question

Calculate the freezing point and boiling point of a solution containing 18.4 g of naphthalene (C10H8) in 114.0 mL of benzene. Benzene has a density of 0.877 g/cm3. (Kf(benzene)=5.12∘C/m.) (Kb(benzene)=2.53∘C/m.)

Answer 1

1) Determine molality of 18.4 g of naphthalene in 114 mL of benzene:

(114.0 mL) x (0.877 g/mL) = 99.978 g = 0.099978 kg

(18.4 g C₁₀H₈) / (128.1705 g C₁₀H₈/mol) / (0.099978kg) = 1.4359 m C₁₀H₈

2) Determine bp elevation:

Δt = i K_b m

Δt = (1) (2.53 °C/m) (1.4359 m)

Δt = 3.63 °C

Normal Boiling Point of Benzene = 80.1^oC

Therefore Boiling point of solution = 80.1 + 3.63 = 83.73^oC

3) Determine fp depression:

Δt = i K_f m

Δt = (1) (5.12 °C/m) (1.4359 m)

Δt = 7.35 °C

Normal Freezing Point of Benzene = 5.5^oC

Therefore Freezing point of solution = 5.5 - 7.35 = -1.85^oC