Question

17. A solution is dissolved by dissolving 0.131 g of substance in 25.4 g of water. The molality is determined
by freezing point depression to be 0.056m. What is the molecular mass of the substance?​

Answer 1

Answer:

The freezing point of a solution that contains 1.00 g of an unknown compound, (A), dissolved in 10.0 g of benzene is found to be 2.07 oC. The freezing point of pure benzene is 5.48 oC. The molal freezing point depression constant of benzene is 5.12 oC/molal. What is the molecular weight of the unknown compound?

Strategy:

Step 1: Calculate the freezing point depression of the solution.

deltaTf = (Freezing point of pure solvent) - (Freezing point of solution)

(5.48 oC) - (2.07 oC) = 3.41 oC

Step 2 : Calculate the molal concentration of the solution using the freezing point depression.

deltaTf = (Kf) (m)

m = (3.41 oC) / (5.12 oC/molal)

m = 0.666 molal

Step 3: Calculate the molecular weight of the unknown using the molal concentration.

m = 0.666 molal = 0.666 mol A / kg benzene

moles A = (0.66 mol A / kg benzene) (0.100 kg benzene) = 6.66 X 10-3 mol A

molecular weight of A = (1.00 g A) / (6.66 X 10-3 mol A)

molecular weigh of A = 150. g/mol