10.00mL 5% acidity vinegar was used titrate with NaOH whose molarity is 0.1030M. Exp. # Concentration of NaOH Initial...

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Chemistry

10.00mL 5% acidity vinegar was used titrate with NaOH whosemolarity is 0.1030M.
Exp. #
Concentration of NaOH
Initial volume of NaOH
Final volume of NaOH
Volume of NaOH used for titration
1
0.1030M
5.50mL
21.65mL
16.15mL
2
0.1030M
3.60mL
19.00mL
15.40mL
3
0.1030M
19.00mL
35.91mL
16.91mL
The vinegar was diluted to make 100.00mL final volume beforetrtration. If the molar mass of acetic acid is 60.05g/mole anddensity is 1.00 g/mL. Calculate the mass percent of acetic acid inthe original vinegar solution.

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4.0 Ratings (808 Votes)

Solution Given data 1000 ml 5 vinegar solution titrated with 01030 M NaOH We have three trials in which volume of the NaOH needed for the titration was calculated So lets first calculate the average of the NaOH solution needed for the titration Balanced reaction See Answer

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Exp. #	Concentration of NaOH	Initial volume of NaOH	Final volume of NaOH	Volume of NaOH used for titration
1	0.1030M	5.50mL	21.65mL	16.15mL
2	0.1030M	3.60mL	19.00mL	15.40mL
3	0.1030M	19.00mL	35.91mL	16.91mL