Question

Freezing points are lowered as a function of the number of moles of solute particles per kilogram of solvent. This is expressed mathematically with the following equation ATt Xkxi (mgolvending) where AT, is the amount by which the freezing point is lowered, Toute is the number of moles of solute, m ening is the mass of the solvent (in kilograms). ke is the freezing point depression constant which is specific to the solvent, and, i is the number of particles per formula unit. Using this equation, find the new freezing point for each of the following solutions for water, kr 1.86 Hint, the new freezing point will be the regular freezing point minus AT a. 3.85 mol NaCl in 4.50 kg of water b. 2.00 g K,PO, in 1.00 kg of water c. 150.0 g of sugar (molar mass - 180.2 g/mol) in 12.50 kg of water

Answer 1

a. molality = moles / mass of water in kg = 3.85 / 4.5 = 0.856 m

Depression in freezing point = molality * K_f = 0.856 * 1.86 = 1.59 ⁰C

Freezing point = -1.59 ⁰C

b. Moles of K₃PO₄ : mass / molar mass = 2 / 212.3 = 9.42 * 10^-3 mol

molality = moles / mass of water = 9.42 * 10^-3 / 1 = 9.42 * 10^-3 m

Depression in freezing point = molality * K_f =9.42 * 10^-3 * 1.86 = 1.75 * 10^-20C

Freezing point : - 0.0175 ⁰C

c. Moles of sugar = mass / molar mass = 150 / 180.2 = 0.832 moles

Moality = moles / mass of water = 0.832 / 12.5 = 0.0666 m

Depression in freezing point = molality * K_f =0.0666 * 1.86 = 0.124 ⁰C

Freezing point : - 0.124 ⁰C