A titration revealed I have .23M acetic acid (HC2H3O2) in a 10.0mL soln.
I was not asked the weigh the sample at any time but our question is:
"The manufacturer of this vinegar claims it is 5% by weight. Ue your results and a density of 1.0g/mL to investigate this claim"
I'm at a loss to determine any results based weight with no weighing involved...I thought the molarity answered any questions about concentration.
Acetic acid 60.05g/mol
No samples weighed at any time
titration began with 10.0mL vinegar and 10.0mL H2O
The Attempt at a Solution
My only guess is acetic acid is 60.05 g/mol at .23M gives me 13.81 grams times .05 gives me .69 g/mL which would be 5% of the molecular weight...?
I'm I getting close?
Thanks as always!