Exercise 2 | Colligative Properties of Solutions

Wines are about 12% ethanol by volume. Assuming ethanol CH3CH2OH is the only nonaqueous constituent, calculate the freezing point of wine.

Data: density of ethanol = 0.79 g.mL-1, freezing point depression constant of ethanol = 1.86 K.kg.mol-1

Let’s determine the freezing point depression due to the ethanol:
 

ΔTf = - i x Kf x m

ΔTf = Tf wine – Tf water

i = van’t Hoff i-factor = number of ions/molecules produced per formula unit

m = solution molality (in mol.kg-1)

Kf = proportionality constant (freezing point depression constant)



12% ethanol by volume: a 100 mL sample of wine contains 12 mL of ethanol and 88 mL of water

methanol = dethanol x Vethanol = 0.79 x 12 = 9.5 g

mwater = dwater x Vwater = 1.00 x 88 = 88 g

m (in mol.kg-1) = nethanol (in mol)mwater (in kg) = methanolMethanol × mwater
m = 9.546.07 × 88 × 10-3 = 2.3 mol.kg-1

 

Freezing point depression:

ΔTf = - i x Kf x m = - 1 x 1.86 x 2.3 = -4.4 K = - 4.4°C

⇒ Tf wine – Tf water =  - 4.4°C

⇒ Tf wine – 0 =  - 4.4°C

⇒ Tf wine = - 4.4°C