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Concentration of NaOH (M): Molar mass of acetic acid: (g/mol) Titrate 5.00 mL of vinegar for each trial. Assume the density of vinegar is 1.00 g/mL. Initial Reading (mL), NaOH buret % error: Final Reading (mL), NaOH buret Total* NaOH volume used in titration (mL) Moles of NaOH used (= moles of acetic acid in 5.00 mL sample) 0.1903 Mass of acetic acid in 5.00 mL sample (g) Experimental % acetic acid Experimental average % acetic acid Deviation Trial 1 0.60 21.60 21 m2 Average Deviation *Total volume = final reading - initial reading Trial 2 5.00 0.30 22.50 22 m2 Trial 3 0.00 22.30 Assume the vinegar is 5.00% acetic acid, as reported on the label. Compare your experimental % acetic acid to the % listed on the label and calculate the % error. |(experimental % - 5.00)| % error = x 100 22.30ml 0.50 22.00 21.3 mu

Concentration of NaOH (M): Molar mass of acetic acid: (g/mol) Titrate 5.00 mL of vinegar for each trial. Assume the density of vinegar is 1.00 g/mL. Initial Reading (mL), NaOH buret % error: Final Reading (mL), NaOH buret Total* NaOH volume used in titration (mL) Moles of NaOH used (= moles of acetic acid in 5.00 mL sample) 0.1903 Mass of acetic acid in 5.00 mL sample (g) Experimental % acetic acid Experimental average % acetic acid Deviation Trial 1 0.60 21.60 21 m2 Average Deviation *Total volume = final reading - initial reading Trial 2 5.00 0.30 22.50 22 m2 Trial 3 0.00 22.30 Assume the vinegar is 5.00% acetic acid, as reported on the label. Compare your experimental % acetic acid to the % listed on the label and calculate the % error. |(experimental % - 5.00)| % error = x 100 22.30ml 0.50 22.00 21.3 mu

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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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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Concentration of NaOH (M): Molar mass of acetic acid: _(g/mol) Titrate 5.00 mL of vinegar for each trial. Assume the density of vinegar is 1.00 g/mL. Initial Reading (mL), NaOH buret % error: Final Reading (mL), NaOH buret Total* NaOH volume used in titration (mL) Moles of NaOH used (= moles of acetic acid in 5.00 mL sample) 0.1903 Mass of acetic acid in 5.00 mL sample (g) Experimental % acetic acid Experimental average % acetic acid Deviation Trial 1 0.60 21.60 21 m2 Average Deviation *Total volume = final reading - initial reading Trial 2 5.00 0.30 22.50 22 тг Trial 3 Assume the vinegar is 5.00% acetic acid, as reported on the label. Compare your experimental % acetic acid to the % listed on the label and calculate the % error. [(experimental % - 5.00)| % error = x 100 0.00 22.30 22.30ml 0.50 22.00 21.3 mu
Transcribed Image Text:Concentration of NaOH (M): Molar mass of acetic acid: _(g/mol) Titrate 5.00 mL of vinegar for each trial. Assume the density of vinegar is 1.00 g/mL. Initial Reading (mL), NaOH buret % error: Final Reading (mL), NaOH buret Total* NaOH volume used in titration (mL) Moles of NaOH used (= moles of acetic acid in 5.00 mL sample) 0.1903 Mass of acetic acid in 5.00 mL sample (g) Experimental % acetic acid Experimental average % acetic acid Deviation Trial 1 0.60 21.60 21 m2 Average Deviation *Total volume = final reading - initial reading Trial 2 5.00 0.30 22.50 22 тг Trial 3 Assume the vinegar is 5.00% acetic acid, as reported on the label. Compare your experimental % acetic acid to the % listed on the label and calculate the % error. [(experimental % - 5.00)| % error = x 100 0.00 22.30 22.30ml 0.50 22.00 21.3 mu
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