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

ID: 955106 • Letter: A

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 6.50 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.

Constants for freezing-point depression and boiling-point elevation calculations at 1 atm:

(°C/m)

Normal freezing

point (°C)

(°C/m)

Normal boiling

point (°C)


*When using positive Kf values, assume that Tf is the absolute value of the change in temperature. If you would prefer to define Tf as "final minus initial" temperature, then Tf will be negative and so you must use negative Kf values. Either way, the freezing point of the solution should be lower than that of the pure solvent.

Solvent Formula Kf value*

(°C/m)

Normal freezing

point (°C)

Kb value

(°C/m)

Normal boiling

point (°C)

water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3 1.22 78.4 carbon
tetrachloride CCl4 29.8 –22.9 5.03 76.8 camphor C10H16O 37.8 176

Explanation / Answer

For a 6.50 gallon = 24605 ml engine coolant and water mixture

with 50/50 mixture

volume of ethylene glycol = 12302.5 ml

mass of ethylene glycol = 12302.5 ml x 1.11 = 13656 g

volume of water = 12302.5 ml

mass of water = 12302.5 x 0.998 = 12278 g

molality of solution = 13656/62.07 x 12.278 = 17.92 m

dTf = -Kf.m = - 1.86 x 17.92 = -33.33 oC

so the freezing point of solution would be -33.33 oC

dTb = Kb.m = 0.512 x 17.92 = 9.17 oC

So the elevation in boiling point of solution = 100 + 9.17 = 109.17 oC

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