Determine the expected freezing point of a calcium chloride solution prepared at room temp. from 2.55kg of calcium chloride and 33.8L of water.

Answer 1

Colligative properites:

Freezing point depression

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 3 for CaCl2 since it produces 3 particles (Ca2+ and 2Cl-)

m = molality = moles solute / kg solvent = 2550 g CaCl2 x 1 mol/111 g = 22.97 mols / 33.8 kg = 0.680 m

K = freezing constant for water = -1.86ºm

Solving for ∆T we have..

∆T = (3)(0.680)(-1.86) = 3.79º

New freezing point = - 3.79 degrees C since normal freezing point is 0ºC