Question

An alcohol, Ethanol (C2H5OH), is present in 1L of wine that is 12% alcohol by volume. The density of the solution is 0.984 g/mL, and the density of pure alcohol is 0.789 g/mL. (C2H5OH = 46.07 g/mol)
Answer these items 1-2 for part 2:
1. What is the weight of the alcohol present in wine?
2. What is the molarity of the solution?

Answer 1

weight of the solution is 0.984 g/mL x 1,000 mL = 984 grams
That 1,000 mL contains 12 mL EtOH/100 mL or 120 mL EtOH in 1,000 mL.
mass = volume x density = 120 mL x 0.789 g/mL = 94.7 g EtOH in 1 L of the wine solution.

M = mols/L. mols EtOH = g/molar mass = 94.7 g/46,1 = 2.05 mols.
That's 2.05 mols/L of solution.