What is the molality of a solution of 31.8 g of ethanol (CH3CH2OH) in 481 mL of water if the density of water is 1.0 g/mL?

Answer 1

The question is asking for molality which is defined as moles of solute per kilogram of solvent.

moles of solute = 31.8 g CH3CH2OH x 1 mol CH3CH2OH / 46.07 g = 0.690 moles solute

kg solvent = 481 ml H2O x 1 g/ml x 1 kg/1000 g = 0.481 kg of solute

Molality = mol solute/kg solvent = 0.690 mol/0.481 kg = 1.44 m