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Question

What is the molality of a solution prepared by dissolving 81.4g of ethylene glycol, HOCH₂CH₂OH, in 1.50 L of water? Assume the density of water is 1.00 g/mL.

Answer 1

molality is defined as moles of solute per kg of solvent

So, we need to find both the moles of solute and the kg of solvent.

moles of solute = 81.4 g x EG x 1 mol EG/62.1 g = 1.31 moles EG

kg of solvent = 1.50 L x 1000 ml/L = 1500 ml x 1 g/ml = 1500 g x 1kg/1000 ml = 1.5 kg

molality = 1.31 mol / 1.5 kg = 0.873 m