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 6.40 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 (HOCH_2CH_2OH), 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.

Explanation / Answer

1 gallon = 3785.41 mL
volume of coolant = 6.4/2 = 3.2 gallons = 3.2*3785.41
volume of coolant = 1.2113 * 10^4 mL
mass of coolant = 1.2113 * 10^4 *1.11 = 1.3445*10^4 g
volume of water = 6.4/2 = 3.2 gallons = 3.2 *3785.41
volume of water = 1.2113 * 10^4 mL
mass of water = 1.2113 * 10^4 * 0.998 = 1.2088*10^4 g
Tb - T0 = Kb * m
Tb - 100 = 0.512 *(1.3445*10^4/62)*(1000/(1.2088*10^4))
Tb = 109.185 C

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