Question

A solution contains 221 grams of calcium chloride in 600 millimeters of water. For the solvent the Kf is 1.86 and Kb is 0.51. What the should the boiling point and the freezing point of the solution be?

Answer 1

first calculate molality of solute

molality = (W /MW) (1000 / mass of solvent in g )

W = 221 g

MW = 110.98 g/mol

mass of solvent = 600 g (density of water = 1.0 g/mL)

molality = (221 / 110.98) (1000 / 600)

molality = 3.32 m

now

CaCl2 ------------> Ca⁺² + 2Cl^-

i = 3 (because 3 ions releasing)

Tf = i x Kf x m

Tf = 3 x 1.86 x 3.32

Tf = 18.52

Tf = 0 - 18.52

Tf = - 18.52⁰C

freezing point = - 18.52⁰C

Tb = i x Kb x m

Tb = 3 x 0.51 x 3.32

Tb = 5.08

Tb = 100 + 5.08

Tb = 105.08⁰C

boiling point = 105.08⁰C