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 5.60 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

1 gallon=3.785 L

Volume of tank=5.60 gallons=5.60*3.785L=21.196L

Volume of each mixed =21.196/2=10.598L

Mass of water taken=density*volume=0.998 g/ml*10.598 *1000 ml=10576.80 g

Mass of ethylene glycol =1.11g/ml*10.598 *1000 ml=11763.78 g

Molality of ethylene glycol, non-volatile solute causing elevation of b.pt needs to be calculated.

Molality of ethylene glycol=moles of ethylene glycol/kg of water

Moles of ethylene glycol=mass/it’s molar mass=11763.78 g/(62.07 g/mol)=189.524 moles

Molality of ethylene glycol=189.524 moles/10576.80 g=189.524/10.576 kg =17.92 moles/kg

∆Tb=elevation of b.pt=i*kb*m

m=molality of solute

kb=boiling-point elevation constant for water=0.512 deg C .kg/mol

i=vant hoff’s factor=1 (solute does not ionize in solution)

∆Tb=1*0.512 deg C .kg/mol * 17.92 moles/kg=9.175 deg C

∆Tb=elevated b.pt of solution-normal b.pt of water at 1 atm=T-100 deg C

9.175 degC= T-100 deg C

T=109.175 deg C (answer)