Question

To what minimum temperature would a radiator be protected if equal volumes of ethylene glycol (permanent antifreeze) (density 1.11 g/mL) and water were mixed. Assume that ethylene glycol does not dissociate and has a formula, C2H6O2.

Answer 1

Assume 2 L of solution

Volume of ethylene glycol = 1 L

Mass of ethylene glycol = volume x density

= 1000 mL x 1.11 g/mL = 1110 g = 1.11 kg

Moles of ethylene glycol = mass/molecular weight

= 1110g / 62g/mol = 17.9 mol

Mass of water = 1000 mL x 1g/ml = 1000g = 1 kg

Freezing point depression

ΔT = i Kf m

i = the number of dissolved particles (Van't Hoff Factor)= 1

Kf = molal freezing point depression constant of the solvent = 1.86 °C/m for water

m = molality = moles of solute / kg of solvent

= 17.9/1 = 17.9 moles /kg

ΔT = 1 x 1.86 x 17.9 = 33.3 °C

Tf° - Tf = 33.3 °C

Tf° = freezing point of pure solvent = 0°C

Tf = freezing point of solution

Tf = - 33.3°C

Below this temperature the solution in the radiator will freeze.

This is the minimum temperature would a radiator be protected.