Answer to Question #314523 in General Chemistry for Nel

Question

To make a solution, 40 g of a nonelectrolyte is dissolved into 300 g of water. The freezing point depression of the solution is measured to be -7°C. Calculate the molar mass of the solute.

Answer 1

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 = van't Hoff factor

K_f = the freezing point depression constant

m = the molality of the solute

The normal freezing point of pure water is 0°C

Therefore, Δt = 0°C − (−7°C) = 7°C

i = 1 (for nonelectrolyte)

K_f = 1.86 °C kg mol¯¹ (for water)

Thus,

m = Δt / (i × K_f)

Molality of solute = (7°C) / (1 × 1.86 °C kg mol¯¹) = 3.7634 mol kg¯¹

Molality of solute = Moles of solute / Kilograms of solvent

Kilograms of solvent = Kilograms of water = (300 g) × (1 kg / 1000 g) = 0.3 kg

Therefore,

Moles of solute = Molality of solute × Kilograms of water

Moles of solute = (3.7634 mol kg¯¹) × (0.3 kg) = 1.129 mol

Moles = Mass / Molar mass

Therefore,

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

Molar mass of solute = (40 g) / (1.129 mol) = 35.43 g/mol

Molar mass of solute = 35.43 g/mol

Answer: The molar mass of the solute is 35.43 g/mol