Chemistry OWLV2

Question

An aqueous antifreeze solution is 22.0% ethylene glycol (

CH;02) by mass. The density of the solution is 1.05

g/cm°. Calculate the molality, molarity, and mole fraction of the ethylene glycol

Answer 1

What does 22.0% by mass mean? It means it has 22.0 g EG per 100 g of solution.

Molar mass EG (ethylene glycol) = 62.1 g / mol

So, to begin, let us assume we have 1 L of the solution (1000 ml).

From density, we can find mass of EG and mass of water

1000 ml x 1.05 g / ml = 1050 g of solution

22% is EG, so 22% x 1050 g = 231 g EG

mass H2O = 1050 g - 231 g = 819 g H2O

We now have all the information we need to find molality, molarity and mole fraction

molality = mols EG / kg H2O

mols EG = 231 g x 1 mol / 62.1 g = 3.72 mols

kg H2O = 819 g H2O x 1 kg / 1000 g = 0.819 kg

molality = 3.72 mol / 0.819 kg = 4.54 m

molartiy = moles EG / L solution

moles EG = 3.72 mols

liters solution = 1

molarity = 3.72 mol / L = 3.72 M

mole fraction = moles EG / moles EG + moles H2O

moles EG = 3.72 mols

moles H2O = 819 g H2O x 1 mol / 18 g = 45.5 mols H2O

mole fraction EG = 3.72 / (3.72 + 45.5) = 0.0756