Question

When 200.0 mg of linalool was added to 100.0 g of camphor the freezing point was lowered by 0.51 oC. What is the molar mass of linalool calculated from its colligative property data? What is the new boiling point of the solution? What is the molarity, weight percent, molality, and ppm concentration of linalool? The density of camphor (and the solution) is 0.992 g , the boiling point is 204 oC, the o mL freezing point is 179.75 C. You will need to find Kf and Kb for camphor.

Answer 1

dTf = -Kf.m

m = moles/kg of solvent = 0.2 g/molar mass x 0.1 kg

0.51 = -(-40 x 0.2/molar mass x 0.1 kg)

molar mass of linalool = 156.863 g/mol

Boiling point of solution = 204 + dTb

dTb = Kb.m

       = 5.95 x 0.2/156.863 x 0.1

      = 0.076 oC

Boiling point of solution = 204 + 0.076 = 204.076 oC

Molarity of linalool = moles/Volume of solution

camphor = 100 g/0.992 = 100.806 ml

concentration of linalool in,

molarity = 0.2/156.863 x 0.101 = 0.013 M

molality = 0.2/156.863 x 0.1 = 0.013 m

weight percent = 0.2 x 100/100.2 = 0.20%

ppm = 200/0.1 = 2000 ppm