The ethyl acetate concentration in a alcoholic solution was determined by diluting a 10.00 mL sample to 100.00 mL. A 20.00 mL portion of diluted solution was refluxed with 40.00 mL of 0.04672 M KOH. Ater cooling, the excess OH2 was back-titrated with 3.41 mL of 0.05042 M H2SO4. Calculate the amount of ethyl acetate (88.11 g mol) in the original sample in grams.

The ethyl acetate concentration in a alcoholic solution was determined by diluting a 10.00 mL sample to 100.00 mL. A 20.00 mL portion of diluted solution was refluxed with 40.00 mL of 0.04672 M KOH. Ater cooling, the excess OH2 was back-titrated with 3.41 mL of 0.05042 M H2SO4. Calculate the amount of ethyl acetate (88.11 g mol) in the original sample in grams.

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter4: Stoichiometry: Quantitative Information About Chemical Reactions

Section4.9: Spectrophotometry

Problem 2.1ACP: Excess KI is added to a 100.0-mL sample of a soft drink that had been contaminated with bleach,...

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The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH. After cooling, the excess OH2 was back-titrated with 3.41 mL of 0.05042 M H2SO4. Calculate the amount of ethyl acetate (88.11 g/mol) in the original sample in grams.
The formaldehyde content of a pesticide was determined by weighing 0.9327 g of the liquid sample into a flask containing 50.0 mL of 0.08624 M NaOH and 50 mL of 3% H2O2. It was heated then cooled. After cooling, the excess base was titrated with 27.8 mL of 0.06325 M H2SO4. Calculate the percentage of HCHO in the sample.
A 1.250 g sample of cheese was subjected to a Kjeldahl analysis to determine the amount of protein. The sample was digested, the nitrogen is oxidized to NH₄⁺, and was then converted to NH₃ with NaOH, and distilled into a collection flask containing 50.00 mL of 0.1050 M HCl. The excess HCl is back titrated with 0.1175 M NaOH, this required 21.65 mL to reach the bromothymol blue end point. A. How many moles of N is present in the cheese sample? B. Report the %N of the cheese sample. C. Report the %protein of the cheese sample, assuming that there are 6.4 grams of protein for every gram of nitrogen
A 1.250 g sample of cheese was subjected to a Kjeldahl analysis to determine the amount of protein. The sample was digested, the nitrogen is oxidized to NH₄⁺, and was then converted to NH₃ with NaOH, and distilled into a collection flask containing 50.00 mL of 0.1050 M HCl. The excess HCl is back titrated with 0.1175 M NaOH, this required 21.65 mL to reach the bromothymol blue end point. What is the reaction for the digestion of the limestone sample A. 2 NaOH + CaCO₃ ⇌ Na₂CO₃ + Ca(OH)₂ B. 2 HCl + CaCO₃ ⇌ CaCl₂ + H₂O + CO₂ C. CaCO₃ ⇌ CaO + CO₂ D. NaOH + HCl ⇌ NaCl + H₂O
The protein in a 1.2846-g sample of an oat cereal is determined by the Kjeldahl procedure for organic nitrogen. The sample is digested with H2SO4, the resulting solution made basic with NaOH, and the NH3 distilled into 50.00 mL of 0.09552 M HCl. The excess HCl is then back titrated using 37.84 mL of 0.05992 M NaOH. Given that the protein in grains averages 17.54% w/w N, report the %w/w protein in the sample of cereal.
A 1.250 g sample of cheese was subjected to a Kjeldahl analysis to determine the amount of protein. The sample was digested, the nitrogen is oxidized to NH₄⁺, and was then converted to NH₃ with NaOH, and distilled into a collection flask containing 50.00 mL of 0.1050 M HCl. The excess HCl is back titrated with 0.1175 M NaOH, this required 21.65 mL to reach the bromothymol blue end point. What is the reaction for the back titration for this analysis? a. NH₄⁺ + NaOH ⇌ NH₃ + H₂O + Na⁺ b. NaOH + HCl ⇌ NaCl + H₂O c. NH₃ + HCl ⇌ NH₄⁺ + Cl⁻ d. N₂ + 3 H₂ ⇌ 2 NH₃ How many moles of N is present in the cheese sample? a. 544 x 10⁻³ b. 706 x 10⁻³ c. 250 x 10⁻³ d. 794 x 10⁻³ Report the %N of the cheese sample. a. 851% b. 884% c. 033% d. 736%
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The amount of protein in a sample of cheese is determined by a Kjeldahl analysis for nitrogen. After digesting a 0.9814-g sample of cheese, the nitrogen is oxidized to NH4+, converted to NH3 with NaOH, and distilled into a collection flask containing 50.00 mL of 0.1047 M HCl. The excess HCl is then back titrated with 0.1183 M NaOH, requiring 22.84 mL to reach the bromothymol blue end point. Report the %w/w protein in the cheese given that there is 6.38 g of protein for every gram of nitrogen in most dairy products.
400 g sample containing chloride ions (MW=35.45 g/mol) was dissolved in 50.0mL of distilled water. A 15.0 mL aliquot was taken, and was further diluted to 50.0 mL using distilled water. The diluted solution was then titrated using a 0.0989 M AgNO3 solution, and it consumed a total of 31.9 mL of the titrant in order to reach the end point. What is the concentration of the undiluted chloride solution? Round off to three (3) significant figures. What is the % (w/v) concentration of the undiluted chloride solution? Round off to two (2) decimal points. What is the % (w/w) of Cl- composition of the sample? Round off to two (2) decimal points.