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.

Answer 1

Cars cooling system ---> 5.5 gallons

1 gallon = 3.78541 Litres

5.5 gallons = 20.82 Litres

50/50 --> engine coolant and water

It means -->10.41 litres of engine coolant and 10.41 litres of water

d = m / v

1.11 = m / 10.41*1000

mass of engine coolant = 11555 gms

mass of water = 0.998*10.41*1000 = 10389 gms

delta T = i*kb*m

i = 1 (non electrolyte)

kb = 0.512

m = molality = moles of solute / mass of solvent in Kg = 11555*1000 / 10389*62 = 17.93

Molarmass of ethylene glycol = 62 gms

delta T = 0.512*17.93 = 9.18 C

Boiling point = boilng poitn of water + delta T = 100 + 9.18 = 109.18 C