Question

Ethylene glycol is used in automobile radiators as an antifreeze. Use the following information to determine the freezing point of an antifreeze solution made by mixing 2.5 L of ethylene glycol with 2.5 L of water. Ethylene glycol is essentially nonvolatile and it does not dissociate in water. Molar mass of ethylene glycol = 62.1 g/mol; density of ethylene glycol = 1.11 g/mL; density of water = 1.00 g/mL; Kf for water = 1.86 degrees Celsius kg/mol

Answer 1

the volume of ethylene glycol added = 2.5 L

The volume of water = 2.5 L

Molar mass of ethylene glycol = 62.1

Density of ehtylene glycol = 1.11

So mass of ethylene glycol taken = Density X volume = 1.11 X 2500mL = 2775 grams

Moles of ethylene glycol taken = Mass / Mol wt = 2775 / 62.1 = 44.69

Density of water = 1g / mL

Mass of water taken =Density x volume = 1 X 2500 mL = 2500 grams

Molality of glycol in water = Moles of glycol / mass of water in Kg = 44.69 / 2.5 Kg = 17.88 molal

From colligative property we know that depression in freezing point of solvent is calculated as

Depression in freezing point = Kf X molality

depression in freezing point = 1.86 X 17.88 = 33.26 ⁰C

So the new freezing point = acutal freezing point - depression in freezing point = -33.26⁰C