Question

Ethylene glycol (C2H4(OH)2) , when dissolved in water, provides the standard ‘anti-freeze’ coolant for water-cooled engines. In order to depress the freezing point of water by 20 °C, how many grams of ethylene glycol would need to be dissolved in 15 kg of pure water? (The molal freezing point depression constant for water Kf = 1.86 K mol-1 kg and the relevant atomic masses are: C = 12g, H = 1g and O = 16g.) Note: ethylene glycol is an organic compound and does NOT break up or dissociate when it dissolves in water. Hint: first calculate the molality (m) of the ethylene glycol solution.

What is the boiling point of this same solution at atmospheric pressure? (The molal boiling point elevation constant for water Kb = 0.52 K mol -1 kg.)

Answer 1

Freezing point depression

Tf - Ts = i x Kf x m

0 - (-20) = 1 x 1.86 x moles of Ethylene glycol / mass of water

20 = 1.86 x moles of Ethylene glycol / 15 kg

moles of Ethylene glycol = 161.29 mol

Mass of Ethylene glycol = moles x molecular weight

= 161.291 mol x 62 g/mol

= 10000 g

molality (m) of the ethylene glycol solution

= moles of Ethylene glycol / mass of water

= 161.291 mol / 15kg

= 10.75

Boiling point elevation

Ts - Tb = i x Kb x m

Ts - 100 = 1 x 0.52 x 10.75

Ts = 105.59 °C = boiling point of this same solution