Question

3. An aqueous solution of 3.63 M glycerol,
C_{3}H_{5}(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**

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