What was the molar concentration of acetic acid in the vinegar?

Expressing concentration of acetic acid in the unit of weight percent?

• Concentration of diluted vinegar = 0.1398mol/L Concentration of acetic acid in undiluted vinegar = 0.02796mol/L Molar Mass (CH3COOH) = 60.052 g/mol Express the concentration of acetic acid in the units of weight percent, i.e: grams of acetic acid per 100 grams of vinegar. Assume that the density of vinegar is 1 g/mL. No idea where to start. Any help will be greatly appreciated. Thanks :)

• Answer:

  0.1398mole/L 1L = 1000g. 0.1398mole x 60.052g/mole = 8.395g. 8.395g / 1000g x 100 = 0.8395% (It doesn't make sense that the molarity of diluted vinegar is higher than the undiluted vinegar, so you may want to check the original problem to make sure you have the right data.)