How to calculate molality with density and molarity?

Calculate the mole fraction, molality and molarity of water in this solution?

• A solution containing 66.0 g of acetone, C3H6O, and 46.0 g of water has a density of 0.926 g/mL. Calculate the mole fraction, molality and molarity of water in this solution. Help!

• Answer:

  (66.0 g acetone) / (58.08 g C3H6O/mol) = 1.1364 mol acetone (46.0 g H2O) / (18.01532 g H2O/mol) = 2.5534 mol H2O (2.5534 mol) / (2.5534 mol + 1.1364 mol) = 0.692 water (mole fraction) (2.5534 mol H2O) / (66.0 g acetone) x (1000 g acetone) = 38.7 m water Take a hypothetical sample of 1.000 L of the mixture: (1000 mL) x (0.926 g/mL) x (46.0 / (46.0 + 66.0)) / (18.01532 g H2O/mol) = 21.1 mol H2O So the molarity is 21.1 M of water.