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:

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

Answer 1

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