Question

A 10.00 ml sample of vinegar, density 1.01g/ml, was diluted to 100.0ml volume. It was found that 25.0 ml of the diluted vinegar required 24.15ml of 0.0976 M NaOH to neutralize it. Calculate the strength of CH3COOH in terms of:

a) molarity
b) grams CH3COOH per litre
c) % Ch3COOH

For part a I used C1V1=C2V2
and found the molarity to be 0.0943M. I don't know if this is correct or not and if it isn't how do I go about doing this question?
I also don't understand how to go about completing parts b and c. I don't want the answer just the explanation on how to do the question.

Thank you for helping!!!

Answers

bobpursley
wouldn't the original vinegar be ten times that calculation, or .943M

grams per liter:
.943=grams/molmass
grams=.943*molmass

percent?
percent=massVinegar/masssolution
= massabove/liter*1liter/1.01*1000

in the last, instead of 10ml, I assumed for calculation 1 liter of the vinegar to determine percent.

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