3. The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.0 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 М КОН: CH3COOC2H5 + OH° ®CH3COO¯ + C2H5OH After cooling, the excess OH was back-titrated with3.41 mL of 0.05042 M H2SO4. Calculate the number of grams of ethyl acetate (88.11 g/mol) per 100 mL of the original sample.
3. The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.0 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 М КОН: CH3COOC2H5 + OH° ®CH3COO¯ + C2H5OH After cooling, the excess OH was back-titrated with3.41 mL of 0.05042 M H2SO4. Calculate the number of grams of ethyl acetate (88.11 g/mol) per 100 mL of the original sample.
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
Chapter1: Chemical Foundations
Section: Chapter Questions
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![3. The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.0
mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672
М КОН:
CH3COOC2H5 + OH° ®CH3COO¯ + C2H5OH
After cooling, the excess OH¯ was back-titrated with3.41 mL of 0.05042 M H2SO4. Calculate the number of grams of
ethyl acetate (88.11 g/mol) per 100 mL of the original sample.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2bafa3d4-8362-4eaf-b0f1-0c492e72839c%2F15eed662-31a2-4850-8a40-35a84bd0e959%2Fnrhdqh_processed.png&w=3840&q=75)
Transcribed Image Text:3. The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.0
mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672
М КОН:
CH3COOC2H5 + OH° ®CH3COO¯ + C2H5OH
After cooling, the excess OH¯ was back-titrated with3.41 mL of 0.05042 M H2SO4. Calculate the number of grams of
ethyl acetate (88.11 g/mol) per 100 mL of the original sample.
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