Question

How many grams of each of the following per kilogram of water in your car radiator are needed to give equal protection against freezing down to -10.0 Degrees Celcius.

A: Methy Alchohol ( CH3OH) Boiling point = 64.6 C
B Ethyele glycol C2H4 )oh)2 Boiling point 197.2 C

C: In spite of the higer cost, what advantage does ethylene glycol possses over metyhly alchol as a winter antifreeze and/or suimmer coolant? Hint conisder the physical properties of each.

Answer 1

find g/kg

We can do this via colligative properties

dTf = -Kf*m

m = molality = mol of solute / kg of solvent

dTf = -10°C, so Kf = -1.86 C/m for water

so

10 = 1.8*molality

molality = 10/1.8 = 5.556 molal

kg of water = 1

mol = 5.556

a)

CH3OH

MW of CH3OH = 32 g/mol

so

mass = 5.556*32

mass = 177.792 g of methanol per kg of water required for 10°C reduction

b)

Ethylene Glycol

MW of EG = 62.07 g/mol

so

mass = 5.556*62.07

mass = 344.86g of methanol per kg of water required for 10°C reduction

C)

Note that Ethylene Glycol is not that volatile as methanol, so you will not lose EG via evaporaiton in hot/warm days, as you will with methanol. Also note that methanol might form an azeotrpe with water.

Since the boiling point is much higher for EG, than methanol, we can achieve high tempearture as well. Recall that the car engine must be cooled, and it has temperautres which approx the 100°C, which is higher than that of BP of methanol, yet EG will not vaporize