Answer to Question #307287 in General Chemistry for Lorena

Question

A solution was made by dissolving 3.75 g of a solute in 108.7 g of acetone . The solution boiled at 56.58° C. The boiling point of pure acetone is 55.95° C, and the Kb =1.71° C/m. What is the molecular weight of the solute?

Answer 1

Solution:

Assume that the solute is a non-electrolyte

The elevation of boiling point of the solvent can be represented using the following equation: 

Δt = i × K_b × m

where:

Δt = the change in boiling point

i = the van't Hoff factor

K_b = the molal boiling-point elevation constant

m = the molality of the solute

 

Δt = T_b(solution) − T_b(pure solvent) = 56.58°C − 55.95°C = 0.63°C

i = 1 (for non-electrolyte)

K_b = 1.71° C/m (for acetone)

Thus,

m = Δt / (i × K_b)

m = (0.63°C) / (1 × 1.71° C/m) = 0.36842 m

Molality of the solute is 0.36842 mol kg⁻¹

Molality = Moles of solute / Kilograms of solvent

Therefore,

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

Moles of the solute = (0.36842 mol kg⁻¹) × (0.1087 kg) = 0.04005 mol

Moles = Mass / Molar mass

Therefore,

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

Molar mass of the solute = (3.75 g) / (0.04005 mol) = 93.63 g/mol

Molar mass of the solute = 93.63 g/mol

Answer: The molecular weight of the solute is 93.63 g/mol