Answer to Question #302013 in Chemistry for mcc

Question

Calculate the molar mass of the solute in a solution of 2.47g of a nonelectrolyte in 100 g of acetic acid (CH₃COOH). The solution freezes at 1.3 °C below the normal freezing point of pure acetic acid. (Kf for CH₃COOH = 61.8°C/m)

Answer 1

The freezing point depression constant of acetic acid (CH₃COOH) is 3.9°C/m (not 61.8°C/m)

Solution:

solute = nonelectrolyte

solvent = acetic acid = CH₃COOH

The lowering (depression) of the freezing point of the solvent can be represented using the following equation: 

Δt = i × K_f × m

where:

Δt = the change in freezing point

i = Vant Hoff factor

K_f = the freezing point depression constant

m = the molality of the solute

Δt = 1.3°C

i = 1 (for nonelectrolyte)

K_f = 3.9°C/m (for CH₃COOH)

m = Δt / (i × K_f)

Therefore,

Molality of the solute = (1.3°C) / (1 × 3.9°C/m) = 0.333 m = 0.333 kg/mol

Molality of the solute = Moles of solute / Kilograms of solvent

Therefore,

Moles of solute = Molality of the solute × Kilograms of solvent

Moles of solute = (0.333 kg/mol) × (0.1 kg) = 0.0333 mol

Moles of solute = Mass of solute / Molar mass of solute

Therefore,

Molar mass of solute = Mass of solute / Moles of solute

Molar mass of solute = (2.47 g) / (0.0333 mol) = 74.17 g/mol

Answer: The molar mass of the solute is 74.17 g/mol