Question

Be sure to answer all parts. How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 7.25 L of water if the coldest winter temperature in your area is -21.0°C? Calculate the boiling point of this water-ethylene glycol mixture. (The density of ethylene glycol is 1.11 g/mL.) What is the volume of antifreeze? L. What is the boiling point of the solution? oC

Answer 1

Volume of antifreeze = 4.58 L

boiling point of the solution = 106 ^oC               (exact boiling point is 105.78 ^oC)

Explanation

Volume of water = 7.25 L

mass of water = (volume of water) * (density of water)

mass of water = (7.25 L) * (1.000 kg/L)

mass of water = 7.25 kg

decrease in freezing point, $\small \Delta$ T_f = 21.0 ^oC

mass of ethylene glycol = ($\small \Delta$T_f * M * m) / (K_f)

where M = molar mass of ethylene glycol = 62.07 g/mol

m = mass of water (kg) = 7.25 kg

K_f = molal freezing point depression constant = 1.86 ^oC.kg/mol

mass of ethylene glycol = [(21.0 ^oC) * (62.07 g/mol) * (7.25 kg)] / (1.86 ^oC.kg/mol)

mass of ethylene glycol = 5080.73 g

volume of ethylene glycol = (mass of ethylene glycol) / (density of ethylene glycol)

volume of ethylene glycol = (5080.73 g) / (1.11 g/mL)

volume of ethylene glycol = 4577.23 mL

volume of ethylene glycol = 4.58 L