Question

A 8.5 mass % aqueous solution of ethylene glycol (HOCH₂CH₂OH) has a density of 1.34 g/mL. Calculate the molarity of the solution.

ANSWER: 1.84 +- 2%

Answer 1

molality = moles of solute / kg solvent

Assume you have exactly 1 kg of solution. 8.5% of that, or 85 grams of the solution is C2H6O2, and 915 g of the solution is water.

Moles C2H6O2 = 85 g / 62.07 g/mol = 1.37 mol C2H6O2

molality = 1.37 mol / 0.6 kg H2O = 2.28 molal
__________________________

Molarity = moles of solute/L of solution
That 1 kg of solution has a volume of:

1000 g / 1.34 g/mL = 746.27 mL = 0.74627 L

Molarity = 1.37 mol / 0.74627 L = 1.84 M