Freezing point depression (mass of additive)

Question

Ethylene glycol (C₂H₆O₂) is used as an additive to the water in your automobile to lower its freezing point. A solution of ethylene glycol in water has a freezing point of -4.60 °C. What mass (in grams) of ethylene glycol must be added to 1000 g of this solution to lower the freezing point to -11.00 °C? (Kf for water is 1.86 °C･kg/mol).

Answer 1

∆T = imK

∆T = change in freezing point

i = van't Hoff factor

m = molality = moles C₂H₆O₂ / kg water

K = freezing constant = 1.86

We need to begin with how we got the solution with a freezing point of -4.60º, since normal freezing point of water is 0º.

∆T = 4.6º

i = 1 since C₂H₆O₂ is a non electrolyte

m = ? = mols C₂H₆O₂ / kg water

K = 1.86

4.60 = (1)(m)(1.86)

m = 2.47 mols C₂H₆O₂ / kg water

molar mass C₂H₆O₂ = 62.1 g / mol

2.47 mols / kg water x 62.1 g / mol = 154 g C₂H₆O₂ / kg water

Since this solution has a mass of 1000 g, the mass of water must be LESS than 1000 g. What is the mass of water present?

g C₂H₆O₂ + g H2O = 1000 g

154 g C₂H₆O₂ + 1000 g H2O = 1154 g

1000 / 1154 = 0.8666

So, 0.8666 x 154 g = 133 g C₂H₆O₂

0.8666 x 1000 g = 867 g H2O

Thus, the original solution with a freezing point of -4.60º contains 133 g C₂H₆O₂ + 867 g H2O

Moving on to the 2nd solution. How much C₂H₆O₂ needs to be added to lower the freezing point from -4.60º to -11.0º?

∆T = imK

∆T = 6.4º

i = 1

m = ?

K = 1.86

6.4 = (1)(m)(1.86)

m = 3.44 mols C₂H₆O₂ / kg H2O

Mass of H2O from above calculation is 867 g = 0.867 kg

3.44 mols C₂H₆O₂ / kg x 0.867 kg = 2.98 mols C₂H₆O₂ needed

2.98 mols C₂H₆O₂ x 62.1 g / mol = 185 g C₂H₆O₂ needs to be added to the original solution