10.00 ml of acetic acid was titratedwith 35.00 ml of 0.10 M NaOH solution,according to the following equation:СНЗСООН + NaOH > CHЗСОONa +H20(M.W of Acetic acid = 60 g*mol, Densityof Vinegar = 1.05 g/ml)Calculate Mass % of acetic acid.1%2%3%4%

Interpretation - To calculate the percentage mass of acetic acid when 10 mL of acetic acid was…

10.00 ml of acetic acid was titrated with 35.00 ml of 0.10 M NaOH solution, according to the following equation: СНЗСООН + NaOH > CHЗСОONa + Н20 (M.W of Acetic acid = 60 g*mol, Density of Vinegar = 1.05 g/ml) %3D Calculate Mass % of acetic acid. 1% 2% 3% 4%

10.00 ml of acetic acid was titrated with 35.00 ml of 0.10 M NaOH solution, according to the following equation: СНЗСООН + NaOH > CHЗСОONa + Н20 (M.W of Acetic acid = 60 g*mol, Density of Vinegar = 1.05 g/ml) %3D Calculate Mass % of acetic acid. 1% 2% 3% 4%

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter4: Reactions In Aqueous Solution

Section: Chapter Questions

Problem 53QAP: The molarity of iodine in solution can be determined by titration with arsenious acid, H3AsO4. The...

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Question

100%

10.00 ml of acetic acid was titrated
with 35.00 ml of 0.10 M NaOH solution,
according to the following equation:
СНЗСООН + NaOH > CHЗСОONa +
H20
(M.W of Acetic acid = 60 g*mol, Density
of Vinegar = 1.05 g/ml)
Calculate Mass % of acetic acid.
1%
2%
3%
4%

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