Question

A 2.950×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.7 mL . The density of water at 20.0∘C is 0.9982 g/mL. Part A Calculate the molality of the glycerol solution.

Answer 1

Answer:

The molality of the glycerol solution is 2.960×10^-2 mol/kg

Explanation:

Number of moles of glycerol = Molarity × volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles

Mass of water = density × volume = 0.9982 g/mL × 998.7 mL = 996.90 g = 996.90/1000 = 0.9969 kg

Molality = number of moles of glycerol/mass of water in kg = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg