Question

How many liters of the antifreeze ethylene glycol would you add to a car radiator containing 6.50 L of water if the coldest winter temperature is -28.0 degrees Celsius? Calculate the boiling point of this water-ethylene glycol mixture. (Then density of ethylene glycol is 1.11g/mL)

Liters of antifreeze _______ L

Boiling point of solution_________ Celsius degrees

Answer 1

1. Freezing point depression = molality x freezing point depression constant of water

28.0 = molality x 1.86

Molality = 15.05 mol/kg

Mass of water = volume x density

= 6.50 L x 1 kg/L = 6.50 kg

Moles of ethylene glycol = molality x mass of water

= 15.05 x 6.50 = 97.85 mol

Mass of ethylene glycol = moles x molar mass

= 97.85 x 62.07 = 6073.5 g

Volume of ethylene glycol = mass/density

= 6073.5/1.11

= 5.47 x 10³ mL = 5.47 L

2. Boiling point elevation = molality x boiling point elevation constant of water

= 15.05 x 0.512 = 7.7 ^oC

Boiling point of solution = boiling point of water + boiling point elevation

= 100.0 + 7.7 = 107.7 ^oC