A bottle of wine contains 11.4% ethanol by volume. The density of ethanol (C2H5OH) is 0.789 g/cm3. Calculate the concentration of ethanol in wine in terms of mass percent and molality.

Answer 1

11.4% v/v means 11.4 mL EtOH/100 mL solution.
Use density to convert 11.4 mL EtOH to grams. mass = v x d = 11.4 x 0.789 = about 9 g (but you need to do it more accurately--all of the other steps of my estimation too).
mols EtOH in 9g = 9/molar mass = about 0.2 mol.
You don't give a density for the solution; therefore, I guess that means we are to assume that the alcohol and water add volumes (which isn't true) but that means 11.4 mL EtOH + 88.6 mL H2O.
The 11.4 mL has a mass of about 9g + 88.6g from the water for a total of about 98g. mass% then is (g EtOH/g soln)*100 = (9/98)*100 = ? % w/w.

molality is mols/kg solvent.
0.2 mols/0.89 kg = ? m