5. If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass percent of CH₂COOH in the vinegar sample? Show your work. Moles of NaOH used 1.5M x 7.5 mL = 11.25 millimole Moles of acetic acid in vinegar 11.25 millimole Concentration of acetic acid = 11.25 millimole/7.0 mL = 1.61 M Mass % of acetic acid= 1.61 M x 60 g/mol (molar mass of acetic acid)= 96.43% ????? I

5. If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass percent of CH₂COOH in the vinegar sample? Show your work. Moles of NaOH used 1.5M x 7.5 mL = 11.25 millimole Moles of acetic acid in vinegar 11.25 millimole Concentration of acetic acid = 11.25 millimole/7.0 mL = 1.61 M Mass % of acetic acid= 1.61 M x 60 g/mol (molar mass of acetic acid)= 96.43% ????? I

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter4: Types Of Chemical Reactions And Solution Stoichiometry

Section: Chapter Questions

Problem 123CP: In most of its ionic compounds, cobalt is either Co(II) or Co(III). One such compound, containing...

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A student had 1.00 L of a 1.00-M acid solution. Much to the surprise of the student, it took 2.00 L of 1.00 M NaOH solution to react completely with the acid. Explain why it took twice as much NaOH to react with all of the acid. In a different experiment, a student had 10.0 mL of 0.020 M HCI. Again, much to the surprise of the student, it took only 5.00 mL of 0.020 M strong base to react completely with the HCl. Explain why it took only half as much strong base to react with all of the HCl.
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