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 6.10 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

Boiling point elevation

dT = Kb x m x i

Volume of solution = 5.30 gallons x 1L/0.26417 gal = 20.06 L

Volume of glycol = 20.06/2 = 10.03 L

Mass of glycol = 1.11 g/mL x 10.03 L x 1000mL/L = 11134.87 g

Moles of glycol = mass/molecular weight

= (11134.87 g) / (62.07 g/mol) = 179.39 mol

Volume of water = 10.03 L

Mass of water = 0.998 g/mL x 10.03 L x 1000mL/L

= 10009.94 g x 1kg/1000g

= 10 kg

Molality = moles of glycol / kg of water

= 179.39 mol / 10 kg

m = 17.94

Boiling point elevation constant Kb = 0.512 °C/m

Van't Hoff factor i = 1

dT = 0.512 x 17.94 x 1

dT = 9.185 °C

Boiling point of solution = 9.185 + boiling point of water

= 9.185 + 100

= 109.185 °C