A solution of ethanol, C2H5OH, in water has a concentration of 2.884 mol L-1. At 20.0 ºC its density is 0.9677?

(a) beginning with 1 L of solution, you know that that contains 2.844 moles of ethanol. Now, the mass of that 1 L of solution can be calculated from the density: 1000 mL X 0.9677 g/mL = 967.7 g of solution.

The mass of the ethanol in that solution is 2.884 mol X 46.07 g/mol = 132.9 grams, which means that the mass of water in that solution is 967.7 g - 132.9 g = 834.8 grams water.

Since molality is defined as moles of solute per kilogram of solvent, the molality of this solution is:

2.884 mol ethanol / 0.8348 kg water = 3.455 molal

(b) the mass percent of a solution is defined as grams of solute / 100 grams solution. So, since the mass of ethanol is 132.9 g and the mass of the solution is 967.7 g, the mass percent of ethanol is:

(132.9 g / 967.7g) X 100 = 13.73% (w/w)