Question

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile\'s engine cooling system. If your car\'s cooling system holds 4.50 gallons, what is the boiling point of the solution? Make the following assumptions in your calculation: at normal filling conditions, the densities of engine coolant and water are 1.11 g/mL and 0.998 g/mL respectively. Assume that the engine coolant is pure ethylene glycol (HOCH2CH2OH), which is non-ionizing and non-volatile, and that the pressure remains constant at 1.00 atm. Also, you\'ll need to look up the boiling-point elevation constant for water.

Answer 1

We will use following formula

ΔTb = Kb X m

Where

ΔTb = elevation in boiling point

Kb = ebulloscopic constant = 0.512 Kg / mole

m = molality

Molality of solute etheylene glycol = Moles of glycol / mass of water in Kg

Total volume = 4.50 gallons,

Volume of ethylene glycol = 2.25 gallons = 8.52 Litres

Mass of ethylene glycol = Density X volume = 1.11g / mL X 8.52 X 10^3 = 9457.2 grams

Moles of ethylene glycol = Mass of ethylene glycol / Molar mass of ethylene glycol = 9457.2 g / 62.07 g / mole

Moles of ethylene glycol = 152.36 moles

Volume of water = 2.25 Gallons = 8.52 Litres

Mass of water = Density X volume = 0.998 g / mL X 8.52 X 10^3 = 8.503 X 10^3 g = 8.503 Kg

Molality = Moles of solute / mass of solvent in Kg = 152.36 moles / 8.503 Kg = 17.92 molal

ΔTb = Kb X m = 0.512 Kg / mole X 17.92 = 9.175

New boiling point = ΔTb + Actual boiling point = 9.175 + 100 ⁰C = 109.175⁰C