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:
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 |
*When using positive Kf values, assume that
ΔTf is the absolute value of the
change in temperature. If you would prefer to define
ΔTf as "final minus initial" temperature, then
ΔTf 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.
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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