Question

How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 5.50 L of water if the coldest winter temperature in your area is −20.0°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?

Answer 1

delta Tf = Kf x m

0 - (- 20) = 1.86 x m

m = 10.75

molality = moles / mass of solvent in kg

10.75 = moles / 5.5 kg

moles = 59.125

moles = mass / molar mass

59.125 = mass / 62

mass of solute = 3665.75 g

density of ethylene - glycol = 1.11 g/mL

volume of ethylene glycol = 3665.75 / 1.11

                                           = 3302

volume of antifreeze = 3.30 L

delta Tb = Kb x m

              = 0.512 x 10.75

Tb - To = 5.504

Tb - 100 = 5.5

Tb = 105.5 oC

boiling point = 105.5 oC