The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 19.00-mL portion of the diluted solutionwas refluxed with 40.00 mL of 0.04601 M KOH:CH3 СОО С2H; + ОН — СH3 COO + C2H; ОНAfter cooling, the excess OH was back-titrated with 3.02 mL of 0.05056 M H2 SO4. Calculate the amount of ethyl acetate (88.11 g/mol) in the original samplein grams.Amount of ethyl acetate =

The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 19.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04601 M KOH: CH; СОО С> Н, + ОН — СH; CОО + С, H; ОН After cooling, the excess OH was back-titrated with 3.02 mL of 0.05056 M H2 SO4. Calculate the amount of ethyl acetate (88.11 g/mol) in the original sample in grams. Amount of ethyl acetate g

The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 19.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04601 M KOH: CH; СОО С> Н, + ОН — СH; CОО + С, H; ОН After cooling, the excess OH was back-titrated with 3.02 mL of 0.05056 M H2 SO4. Calculate the amount of ethyl acetate (88.11 g/mol) in the original sample in grams. Amount of ethyl acetate g

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.20QAP

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Calculate the percentage CH3COOH in a sample of vinegar from the following data.Sample = 15.00 g, NaOH used = 43.00 ml ; 0.600 N H2SO4 used for back titration =0.250 ml ; 1.00ml NaOH is equivalent to 0.0315 g H2C2O4. 2H2O.
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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