Answer to Question #302007 in Chemistry for mcc

Question

What are the freezing point and boiling point of a solution that contains 5 g of glycerol (C₃H₈O₃) and 20 ml of water. 

Answer 1

Solution:

solute = glycerol = C₃H₈O₃

solvent = water = H₂O

The molar mass of glycerol (C₃H₈O₃) is 92.09 g/mol

Therefore,

Moles of glycerol = (5 g) × (1 mol / 92.09 g) = 0.0543 mol C₃H₈O₃

The density of water is around approximately 1 g/mL

Therefore,

Mass of water = (20 mL) × (1 g / 1 mL) = 20 g H₂O

Convert g to kg:

Mass of water = (20 g) × (1 kg / 1000 g) = 0.02 kg H₂O

Molality = Moles of solute / Kilograms of solvent

Therefore,

Molality of glycerol = Moles of glycerol / Kilograms of water

Molality of glycerol = (0.0543 mol) / (0.02 kg) = 2.715 mol/kg = 2.715 m

The freezing point or boiling point of the solvent can be represented using the following equation: 

Δt = i × K × m

where:

Δt = the change in the freezing/boiling point

i = Vant Hoff factor

K = the freezing point or boiling point depression constant

m = the molality of the solute

The freezing point of pure water is 0°C.

i = 1 (for glycerol)

K_f = 1.86°C/m (for water)

Therefore,

Δt = (1 × 1.86°C/m × 2.715 m) = 5.05°C

The freezing point = T_f = 0°C − 5.05°C = −5.05°C

The freezing point of the solution is −5.05°C

The boiling point of pure water is 100°C.

i = 1 (for glycerol)

K_b = 0.512°C/m (for water)

Therefore,

Δt = (1 × 0.512°C/m × 2.715 m) = 1.39°C

The boiling point = T_f = 100°C + 1.39°C = 101.39°C

The boiling point of the solution is 101.39°C

Answers:

The freezing point of the solution is −5.05°C

The boiling point of the solution is 101.39°C