Chem II Lab- Titration of Vinegar with NaOH

[ChocolateCoke]

Homework Statement

"Calculate the grams of acetic acid per liter of vinegar"

Homework Equations

Previous problem stated the multiply the average molarity of diluted vinegar times 10= 2.7 M

Vinegar (acetic acid)= H(C2H3O2) molecular weight= 60.06 g

The Attempt at a Solution

moles= M x L
2.7M x 1L = 2.7 moles
2.7 moles x 60.06 g= 162.16 g

This seems way too high. and i wasn't given anything really and it is driving me insane!

Thank you in advance for anyone who helps!

 

[symbolipoint]

To the extent that you described your situation, the work and the result look good.

Note that you labeled the topic, "...Titration of Vinegar...". We may assume you were doing calculations to prepare to perform a titration.

 

[Borek]

2.7M acetic acid is about 16% - strong for vinegar, but not impossible.

In Germany in grocery stores you can buy Essigessenz which is basically just 70% acetic acid.