Question

Ethylene glycol (molar mass = 62 g mol¯ 1) is a common automobile antifreeze. Calculate the freezing point of a

solution containing 12.4g of this substance in 100 g of water. Would it be advisable to keep this substance in the car

radiator during summer?

Given : Kffor water = 1.86K kg/mol

Kbfor water = 0.512K kg/mol

Answer 1

Freezing point depression ΔT_f = K_f *b * i

i = 1 for ethylene glycol

b = moles of solute / kg of solvent

1 mole of Ethylene glycol = 62g

Therefore, 12.4g constitutes 12.4/62 = 0.2 mole

Solvent = 100g = 0.1kg

Hence, b = 0.2/0.1 = 2 moles/kg

ΔT_f = 1.86 * 2 Kelvins

ΔT_f = 3.72 Kelvins

Difference in freezing point = 3.72K (depression)

We know freezing point of pure water is 0 deg C = 273K

Therefore freezing point of solution =273 – 3.6 = 269.4K = -3.72 deg C

(since temp in K = 273 + temp in C)

Boiling point elevation ΔT_b= K_b *b * i

= 0.512 * 2 = 1.024K

Boiling point of pure water = 100 deg C = 373K

Therefore, boiling point of solution = 373 + 1.024 = 374.024K

= 101.024 deg C

However, the boiling point for aqueous ethylene glycol increases with increasing ethylene glycol percentage. Thus, the use of ethylene glycol not only depresses the freezing point, but also increases the boiling point so it is good top keep in radiator during summer.