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 4.70 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.

Answer 1

Convert 4.70 gallons to ml

4.70 gallons = 177914ml

1/2 of that will be ethylene glycol; 1/2 will be water. 177914/2 = 88957 ml

Using density, convert ml H₂O to grams and ml ethylene glycol to grams

88957ml x 0.998g/ml = 88779.086 g water = 88.779 Kg water

88957ml x 1.11g/ml = 98 742.27g ethylene glycol

Convert grams glycol to moles.

Molar mass of ethylene glycol = 62.08g/mole

Moles of ethylene glycol = grams/molar mass

   = 98742.27g/62.08g/mole

                                           = 1590.56 moles

Convert moles glycol to molality. m = moles/kg solvent

So, there are 1590.56 moles in 88.779 Kg water but, we need the moles in 1Kg of water:

1590.56 moles / 88.779 Kg = 17.915 moles/kg water, so ethylene glycol is 17.915 molal

Then ΔT= Kb*m

The boiling-point elevation constant Kb for water is = 0.52⁰C/molal

ΔT = 0.52⁰C x 17.915 molal = 9.3⁰C

The boiling point should be 100⁰ C+ 9.3⁰C = 109.3⁰C