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 5.50 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.

Solvent Formula Kf value k Normal freezing Kb value Normal boiling (°C/m) point (OC) (OC/m) point (OC) 0.512 100.00 H20 1.86 0.00 water 5.49 2.53 80.1 C6H6 5.12 benzene cyclohexane C6H12 20.8 6.59 2.92 80.7 -117.3 78.4 ethanol C2H60 1.99 1.22 CCI 22.9 Carbon 29.8 5.03 76.8 tetrachloride C10H160 37.8 176 Camphor When using positive Ke values assume that ATe is the absolute value of the change in

Explanation / Answer

50/50 ----> Water and Coolant

VOlume = 5.5 gallons

1 gallon = 3.78 litres

5.5 gallons = 20.79 litres

Volume of coolant = 20.79/2 = 10.395 L

Volume of water = 10.395 L

d = mass / Volume

Mass of ethylene glycol = 1.11 *[10.395*1000] = 11538.45 gms

Mass of water = 0.998*10.395*1000 = 10374.21 gms

delta Tb = Kb*m

delta Tb = elevation in boiling point ; Kb = boiling point constant

m = molality = moles of solute / Mass of solvent in Kg

solute = ethylene glycol and solvent is water

Molar mass of ethylene glycol = 62 gms/mol

delta Tb = 0.512 *11538.45 *1000 / [10374.21*62] = 9.184 C

Boiling point of solution = 100C + 9.184 C = 109.184 C

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