A 50/50 blend of engine coolant and water (by volume) is usually used in an auto

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 5.30 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 (HOCH2CHzOH), 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 Number 1 Tools x 10

Explanation / Answer

Tb = i. Kb. m

Where: Tb =elevation in the boiling point = Tf – Ti

Ti = boiling point of water = 100oC

i= Vant hoff factor

Kb = boiling point elevation constant.

m = molality

i = 1 as ethylene glycol is non-electrolyte

Given to assume coolant is pure ethylene glycol so,

Volume of coolant => 5.30 gallons = 20.0627 Litres = 20062.7 ml

coolant = water : EG = 50:50

Water = 20062.7/2 = 10031.35 ml

Density = 0.998 g/ml

Mass of water = 10031.35 ml x 0.998 g/ml = 10011.28 g..............(1)

Volume of ethylene glycol = 10031.35 ml

Density of ethylene glycol = 1.11 g/ml

D = M/V

Mass = D X V

Mass = 1.11 g/ml x 10031.35 ml = 11134.79 g ...................(2)

Molar mass of ethylene glycol = 62.07 g/mol

No. of moles = 11134.79/62.07 = 179.39 moles.................(3)

Solvent = 10011.28g = 10.01 kg

Molality = 179.39 moles/10.01 kg = 17.92 m .................(4)

Kb = 0.512 oC./m

Tb = i. Kb. m

Tb = 0.512 oC./m. X 17.92 m

Tb = 9.17 oC

Tb = Tf – Ti

Tf = Ti + Tb

Tf = (100 + 9.17) oC = 109.17 oC

Final boiling point of coolant = 109.17 oC

Elevation in boiling point Tb = 9.17 oC

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