Answer to Question #301501 in General Chemistry for Stephany

Question

Ethylene glycol C2H6, is a common automobile antifreeze agent. Calculate the freezing point of solution containing 651 grams of EG in 2505 grams of water. The molar mass of EG is 62.01 g/mol (Kf=1.86° C/m) and (Kb= 0.52 ° C/m for water)

Show your solution.

Answer 1

Depression in any solute is known as the colligative property.

Here, ΔTf = kf*m m = molality which is equal to moles of solute per kg solvent

Moles of ethylene glycol = 12.4/62

Molality = 0.2/0.1=2

By substituting we get,

ΔTf = 1.86*2 = 3.72K

Difference in freeezing point = 3.72K ie Depression

Boiling point elevation = ΔTb = Kb*b*i

=0.512*2 = 1.024K

Boiling point of solution = 373+1.024= 374.024k= 101.024deg C

The boiling point increases with increase in ethylene glycol percentage.

The use of ethylene glycol increases boiling point and depresses the freezing point.

It act as a anti freezing agent also and can be keep this in car radiator during summer