Given that the density of vinegar is 1.049 g/mL, convert the molarity of the acetic acid solution found in calculation 5 into a mass percentage of acetic acid.(calculation from number 5 is Molarity of acetic acid which is 2.251)
M = 2.251 moles/L of solution.
grams CH3COOH = moles x molar mass = 2.251 x 60 = 135.06
mass of the solution:
mass = volume x density = 1,000 mL x 1.049 g/mL = 1,049 grams
% CH3COOH w/w = (135.06 g/1049 g)*100 = ?