Answer to Question #185991 in Chemistry for Jasmine Jackson

Question

If 0.231 mol of potassium fluoride (KF) is dissolved in 50.3 g of water, how much will the freezing point be lowered? (17.1°C)

Answer 1

Solution:

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

ΔT_f = i × K_f × m (for electrolytes)

where:

i = the number of dissolved particles (van’t Hoff factor).

m = molality = moles of solute per kilogram of solvent.

K_f is the molal freezing point depression constant of the solvent (1.86°C/m for water).

The van't Hoff factor is the number of particles (ions) formed by the compound in the solution.

The dissociation of KCl is given by,

KCl → K⁺ + Cl^-

Therefore, van’t Hoff factor is 2 for KCl.

Molality of KF solution = Moles of KF / Kilograms of H₂O

Molality of KF solution = (0.231 mol) / (0.0503 kg) = 4.5924 mol/kg = 4.5924 m

Thus,

ΔT_f = (2) × (1.86°C/m) × (4.5924 m) = 17.08°C = 17.1°C

ΔT_f = 17.1°C

Answer: The freezing point will be lowered by 17.1°C