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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
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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
Transcribed Image Text: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
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