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!!!

Answer 1

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.