Molar Mass determination by depresson of the freezing point?

Question

1. An aqueous solution of sucrose was prepared by dissolving 34.58 g of C12H22O11 (M.M. = 342.34) in 100.0 mL of distilled water.

a. How many moles of sucrose were dissolved in water?

b. What is the solvent used to prepare the solution?

c. What is the mass (kg) of the solvent?

d. Calculate the molality of the sucrose solution.

Answer 1

a.  (34.58 g)/(342.34 g/mol) ≈

0.1010107 mol ≈ 0.101 mol

b. water

c. 0.100 kg

d. 1.01 mol/kg

Answer 2

Actually

a. (34.58 g)/(342.34 g/mol) ≈

0.1010107 mol ≈ 0.101 mol

b. water

c. 0.100 kg

d. 1.01 mol/kg