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.
Chapter7: Solutions And Colloids
Section: Chapter Questions
Problem 7.34E
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