Answer to Question #328639 in Chemistry for Max

Question

Camphor (C10H16O) has a molal freezing point depression constant of 5.95°C/m. If you dissolve 10.0 g of dimethyl ether (C2H6O) into 3.00 kg of camphor, what will be the change in the freezing point of camphor?

Answer 1

Solution:

solute = dimethyl ether (C₂H₆O)

solvent = camphor (C₁₀H₁₆O)

The equation describing the change in freezing point from pure solvent to solution is:

Δt = i × K_f × m

where:

Δt = the change in freezing point (°C)

i = the van't Hoff factor

K_f = the molal freezing point depression constant (°C m⁻¹)

m = the molality of the solute (m)

The molar mass of dimethyl ether (C₂H₆O) is 46.07 g mol⁻¹

Therefore,

Moles of C₂H₆O = (10.0 g C₂H₆O) × (1 mol C₂H₆O / 46.07 g C₂H₆O) = 0.217 mol C₂H₆O

Molality = Moles of solute / Kilograms of solvent

Therefore,

Molality of C₂H₆O solution = (0.217 mol) / (3.00 kg) = 0.0723 mol/kg = 0.0723 m

 

i = 1 (for nonelectrolyte - dimethyl ether)

K_f = 5.95°C m⁻¹ (for camphor)

Thus,

Δt = (1) × (5.95°C m⁻¹) × (0.0723 m) = 0.43°C

Δt = 0.43°C

 

Answer: The change in the freezing point of camphor will be 0.43°C