# Hon many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a?

Hon numerous liters for the antifreeze ethylene glycol [CH2(OH)CH2(OH)] can you enhance a motor vehicle radiator containing 6.50 L of liquid if coldest cold temperatures heat in your town is

The freezing point despair is
delta Tf = typical FP of H2O – desired FP of H2O = 0 C – (-20.5 C) = 20.5 C
delta Tf = i x Kf x molality EG
i = 1 for nonelectrolytes like EG
Kf = molal freezing point continual for H2O = 1.86 C/m
6.50 L H2O x (1000 mL H2) / 1 L H2O) x (0.997 g H2O / 1 mL H2O) = 6480 g H2O = 6.48 kg H2O
delta Tf = i x Kf x molality EG
20.5 = (1)(1.86)(molality EG)
molality EG = 11.0 molal
molality EG = moles EG / kg H2O
11.0 = moles EG / 6.48
moles EG = 71.3
71.3 moles EG x (62.1 g EG / 1 mole EG) = 4430 g EG
4430 g EG x (1 mL EG / 1.11 g EG) = 3990 mL EG = 3.99 L
Examine my mathematics.
delta Kb = i x Kb x molality EG = (1)(0.512)(11.0) = 5.63 C
brand-new BP = typical BP + delta Kb = 100.00 + 5.63 = 105.63 C.

Antifreeze Ethylene Glycol

Desired freezing point despair = 20.5° C
Kg of solvent (H2O) = 6.50 kg
Freezing point despair = Kb x m
Kb = molal freezing-point despair constant
20.5 = 1.86 x molality of ethylene glycol
20.5 = 1.86 x moles of ethylene glycol / kg of solvent
kg of solvent = 6.50
20.5 = 1.86 x moles of ethylene glycol / 6.50 kg
20.5 x 6.50 / 1.86 = moles of ethylene glycol = 71.6
71.6 mole x 62.1 g/mole = 4449 g (4.45 kg)
1 L = 1000 mL x 1.11 g/mL = 1100 g = 1.1 kg
4.45 kg / 1.10 kg/L = 4.04 L
Buy 4 L of ethylene glycol

(-20.5° C) / (-1.86 °C/m) = 11.022 m
(11.022 mol / 1000 g H2O) x (6500 g H2O) = 71.643 mol ethylene glycol
(71.643 mol C2H6O2) x (62.0678 g C2H6O2/mol) / (1.11 g/mL) = 4006 mL = 4.01 L ethylene glycol
(11.022 m) x (0.512 °C/m) = 5.64 °C modification
100°C + 5.64 °C = 105.6 °C