Question

Calculate the freezing point and boiling point of a solution containing 18.4 g of naphthalene (C10H8)...

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.)

Homework Answers

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 Kb m

Δt = (1) (2.53 °C/m) (1.4359 m)

Δt = 3.63 °C

Normal Boiling Point of Benzene = 80.1oC

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

3) Determine fp depression:

Δt = i Kf m

Δt = (1) (5.12 °C/m) (1.4359 m)

Δt = 7.35 °C

Normal Freezing Point of Benzene = 5.5oC

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

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