50/50 blend of engine coolant and water (by volume) is usually used in an automo
ID: 715566 • Letter: 5
Question
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 6.00 gallons, what is the boiling point of the solution? Make the following assumptions in your calculation: at normal filling conditions, the densities of engine oolant 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. NumberExplanation / Answer
Solution :-
Total volume of engine coolant = 6.00 gallons
(50/50)blend of ethylene glycol and water
Lets convert gallons to ml
6.00 gal * 3785.41 ml / 1 gal = 22712 ml
Volume of ethylene glycol = 22712 ml / 2 = 11356 ml
Volume of water = 11356 ml
Converting volume to mass
Mass = volume x density
Mass of ethylene glycol (solute)= 11356 ml * 1.11 g/ml = 12605 g
Mass of water (solvent)= 11356 ml * 0.998 g/ml = 11333 g
Converting grams to kg
11333 g * 1 kg / 1000 g = 11.333 kg
Lets calculate moles of ethylene glycol
Molar mass of ethylene glycol (OHCH2CH2OH) = 62.07 g/mol
Moles of ethylene glycol = mass / molar mass
= 12605 g / 62.07 g per mol
= 203.4 mol
Molality = moles of solute / kg solvent
= 203.4 mol / 11.333 kg
= 17.95 m
Calculating the change in the boiling point (delta Tb)
Molal boiling point constant of water = 0.512 oC/m
Delta Tb= Kb* m
= 0.512 oC/m * 17.95 m
= 9.19 oC
Boiling point of coolant = boiling point of pure water + delta Tb
= 100 oC + 9.19 oC
= 109.2 oC
Therefore boiling point of coolant is 109.2 oC
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