Question

two aqueous solutions are prepared. Solution A is prepared by dissolving 50 grams of ethylene glycol (62.07 g/mol) into 1 kg of water. Solution B is prepared by dissolving 50 grams of propylene glycol (76.09 g/mol) into 1 kg of water. Calculate the freezing points for each.

Answer 1

Solution A : ethylene glycol

Given that Solution A is prepared by dissolving 50 grams of ethylene glycol (62.07 g/mol) into 1 kg of water.

molality = (mass / molar mass) x (1/ mass of solvent in kg)

= (50 g / 62.07 g/mol) x (1/ 1 kg)

= 0.805 mol/kg

molality of ethylene glycol = 0.805 mol/kg

molal freezing point depression constant of water, Kf = 1.86 oC.kg/mol

Freezing point depression (ΔT) = Kf m

= 1.86 oC.kg/mol x 0.805 mol/kg

= 1.5 oC

ΔT = 1.5 oC

Therefore, Freezing point depression (T) of ethylene glycol = 1.5 oC

Freezing point depression (ΔT) = Freezing point of water - Freezing point of ethylene glycol solution

Then , Freezing point of ethylene glycol solution

= Freezing point of water - Freezing point depression (T)

= 0oC - 1.5 oC

= - 1.5 oC

Freezing point of ethylene glycol solution = - 1.5 oC

Solution B : propylene glycol

Given that Solution B is prepared by dissolving 50 grams of propylene glycol (76.09 g/mol) into 1 kg of water.

molality = (mass / molar mass) x (1/ mass of solvent in kg)

= (50 g / 76.09 g/mol ) x (1/ 1 kg)

= 0.657 mol/kg

molality of propylene glycol = 0.657 mol/kg

molal freezing point depression constant of water, Kf = 1.86 oC.kg/mol

Freezing point depression (ΔT) = Kf m

= 1.86 oC.kg/mol x 0.657 mol/kg

= 1.22 oC

ΔT = 1.22 oC

Therefore, Freezing point depression (T) of propylene glycol = 1.22 oC

Freezing point depression (ΔT) = Freezing point of water - Freezing point of propylene glycol solution

Then , Freezing point of propylene glycol solution

= Freezing point of water - Freezing point depression (T)

= 0oC - 1.22 oC

= - 1.22 oC

Freezing point of propylene glycol solution = - 1.22 oC