Question

An aqueous antifreeze solution is 70.0% ethylene glycol (c2h6o2) by mass. the density of the solution is 1.75 g/cm3. calculate the molality, molarity, and mole fraction of the ethylene glycol.

Answer 1

Molarity = moles of solute / kg solvent

Assume you have exactly 1 kg of solution. 70% of that, or 700 grams of the solution is C2H6O2, and 300 g of the solution is water.

Moles C2H6O2 = 700 g / 62.07 g/mol = 11.28 mol C2H6O2

molarity = 11.28 mol / 0.3 kg H2O = 37.59 molal

__________________________

Molarity = moles of solute/L of solution

there is 1 kg of solution has a volume of:

1000 g / 1.05 g/mL = 952 mL = 0.952 L

Molarity = 11.28 mol / 0.952 L = 11.85 M

___________________________

Mole fraction ethylene glycol = moles ethylene glycol / (moles ethylene glycol + moles H2O)

moles ethylene glycol = 11.28

moles H2O = 300 g / 18.0 g/mol = 16.67 mol H2O

mole fraction = 11.28 /( 11.28 + 16.67) = 0.404