Exercise 2 | Colligative Properties of Solutions

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

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

m_ethanol = d_ethanol x V_ethanol = 0.79 x 12 = 9.5 g

m_water = d_water x V_water = 1.00 x 88 = 88 g

m (in mol.kg^-1) = $\frac{{\mathrm{n}}_{\mathrm{ethanol}} (\mathrm{in} \mathrm{mol})}{{\mathrm{m}}_{\mathrm{water}} (\mathrm{in} \mathrm{kg})}$ = $\frac{{\mathrm{m}}_{\mathrm{ethanol}}}{{\mathrm{M}}_{\mathrm{ethanol}} \times {\mathrm{m}}_{\mathrm{water}}}$
m = $\frac{9.5}{46.07 \times 88 \times {10}^{-3}}$ = 2.3 mol.kg^-1

Freezing point depression:

ΔT_f = - i x K_f x m = - 1 x 1.86 x 2.3 = -4.4 K = - 4.4°C

⇒ T_{f wine} – T_{f water} =  - 4.4°C

⇒ T_{f wine} – 0 =  - 4.4°C

⇒ T_{f wine} = - 4.4°C