Question

How many grams of each of the following per 1000g of water in your car radiator are needed to give equal protection against freezing down to -10.0°C? (a) Methyl alcohol (CH3OH) b.p is 65°C, (b) ethyl alcohol (C2H5OH) b.p is 78°C, (c) ethylene glycol (C2H4(OH)2) b.p is 197°C. In spite of higher cost, what advantages does ethylene glycol have over the other alcohols as an all-winter antifreeze?

Answer 1

Expression for depression in freezing point can be given as follows:

ΔT_f = K_f . b . i

Where, ΔT_f = depression in freezing point = 10°C = 10K

K_f = cryoscopic constant for water = 1.853 K. kg/mol

b = molality

i = van’t Hoff factor = 1 for molecules that do not dissociate.

Thus,

10K = (1.853 K. kg/mol) x b x 1

b = 10 / 1.853

b = 5.40 mol/kg of solution

The number of moles of methyl alcohol, ethyl alcohol and ethylene glycol can be calculated as follows:

1. Methyl alcohol has molar mass of 32 g/mol. Thus, weight of 5.40 molal methylalcohol can be calculated as follows:
   weight of methyl alcohol = 5.40 x 32 = 172.8 g of methyl alcohol in 1000g solution
   

2. Ethyl alcohol has molar mass of 46 g/mol. Thus, weight of 5.40 molal ethylalcohol can be calculated as follows:
   weight of ethyl alcohol = 5.40 x 46 = 248.4 g of ethyl alcohol in 1000g solution
   

3. Ethylene glycol has molar mass of 62 g/mol. Thus, weight of 5.40 molal ethylene glycol can be calculated as follows:
   weight of ethylene glycol = 5.40 x 62 = 334.8 g of ethylene glycol in 1000g solution

Therefore, to get protection against freezing to -10.0°C, car radiator needs 172.8 g of methyl alcohol in 1000g solution, 248.4 g of ethyl alcohol in 1000g solution, 334.8 g of ethylene glycol in 1000g solution.

Despite of higher cost ethylene glycol has advantages over other alcohols due to its high boiling point. Due to high temperature in radiator, methyl alcohol and ethyl alcohol will evaporate fast from the solution of coolant. Ethylene glycol, due to its very high boiling point (197°C), will not evaporate easily.