Question

3. An aqueous solution of 3.63 M glycerol, C₃H₅(OH)₃, has a density of 1.185 g/mol.

a) Calculate molar concentration of the solution.

b) Find the weight percent of glycerol in the solution.

c) What is the mole fraction of glycerol in the solution?

Answer 1

a) Molar concentration of the solution

=> molarity * density

=> 3.63 * 1.185

=> 4.30 M

b) Number of moles of Glycerol in 1L of solution = 3.63 moles

Molecular weight of Glycerol = 128.13 g/mol

Mass of Glycerol = 3.63 * 128.13 = 465.119 gms

Mass of solution = mass of solute + mass of solution = 465 + 1000 = 1465

Percentage by mass = 465 * 100/1465 = 31.70%

c)

Mole fraction of glycerol in solution = 3.63 moles

Weight of water = 1000 g

Number of moles = 1000/18 = 55.55 moles

Mole fraction of Glycerol = 3.63/(3.63+55.55) = 0.0613