Question

Calculate the freezing point and boiling point of a solution containing 16.0 g of naphthalene (C10H8) in 109.0 mL of benzene. Benzene has a density of 0.877 g/cm3.

Part A

Calculate the freezing point of a solution. (Kf(benzene)=5.12∘C/m.)

Part B

Calculate the boiling point of a solution. (Kb(benzene)=2.53∘C/m.)

Answer 1

Sol :- Given mass of naphthalene = 13.4 g

Volume of benzene = 110.0 mL = 110.0 cm³

Benzene has a density = 0.877 g/cm³

So, Mass of benzene (solvent ) = Density x volume = 110.0 cm³ x 0.877 g/cm³ = 96.47 g = 0.09647 kg

We know, Molality = Number of moles of solute / Mass of solvent in kg ...........(1)

Number of moles of solute i.e. naphthalene = mass / Gram molar mass = 13.4 g / 128.17052 g/mol = 0.10455 mol

So,

Molality = 0.10455 mol / 0.09647 kg = 1.084 mol/kg or 1.084 m

We know that,

(a). Depression in freezing point = ΔT_f = T_f⁰ - T_f = i x k_f x m

here, T_f⁰ = Freezing point of pure solvent i.e. benzene = 5.5 °C

T_f = Freezing point of solution = ?  

k_f = molal freezing point constant of solvent = 5.12°C/m

i = vant's hoff factor of naphthalene = 1 ( for non-electrolyte)

So,

T_f⁰ - T_f = i x k_f x m

5.5 °C - T_f = 1 x 5.12 ⁰C/m x 1.084 m

- T_f = 5.55008 ⁰C - 5.5 ⁰C

T_f = - 0.050 ⁰C

Hence, freezing point of solution = - 0.050 ⁰C

(b). Elevation in boiling point = ΔT_b = T_b - T⁰_b= i x k_f x m

here, T_b⁰= boiling point of pure solvent i.e. benzene = 80.1 °C

T_b= boiling point of solution = ?  

k_b = molal boiling point constant of solvent = 2.53 °C/m

i = vant's hoff factor of naphthalene = 1 ( for non-electrolyte)

So,

T_b - T⁰_b= i x k_f x m

T_b - 80.1 °C = 1 x 2.53 ⁰C/m x 1.084 m

T_b = 2.74 ⁰C  + 80.1  ⁰C

T_b = 82.4 ⁰C

Hence, boiling point of solution = 82.4 ⁰C