Calculate the freezing point and boiling point of an antifreeze solution that is 47.3% by mass ethylene glycol (HOCH2CH2OH) in water. Ethylene glycol is a nonelectrolyte.

Question

Calculate the freezing point and boiling point of an antifreeze solution that is 47.3% by mass ethylene glycol (HOCH₂CH₂OH) in water. Ethylene glycol is a nonelectrolyte.

Freezing point (in °C): _______

Boiling point (in °C): ________

Answer 1

Freezing point depression/ boiling point elevation: ∆T = imK

∆T = change in freezing or boiling point = ?

i = van't Hoff factor = 1 for ethylene glycol (a non electrolyte)

m = molality = moles ethylene glycol per kg water (see below)

K = freezing or boiling point constant: Kf = 1.86; Kb = 0.512 º/m

Finding m (molality): For 1 kg of solution, 47.3% is ethylene glycol. Thus 473 g of ethylene glycol are present.

Mass of water = 1000 g - 473 g = 527 g = 0.527 kg water

moles ethylen glycol = 473 g x 1 mol/ 62.07 g = 7.62 moles

m = 7.62 mol/0.527 kg = 14.5 molal

Freezing point:

∆T = (1)(14.5 m)(1.86 º/m) = 27.0º

Freezing point = -27.0ºC

Boiling point:

∆T = (1)(14.5 m)(0.512º/m) = 7.42º

Boiling point = 107ºC