0.9583 g of sample was dissolved in 80 mL of xylene and extracted with three 15 mL quantities of water. The combined aqueous extracts were diluted to 50.0 mL and then shaken with 10 mL toluene. The toluene was then discarded. The absorbance of the aqueous solution when measured in a 1cm path length cell was 0.134 at 278 nm.Calculate the free chlorampenicol content of the sample in ppm, given A(1%, 1cm) = 298 at 278 nm.

0.9583 g of sample was dissolved in 80 mL of xylene and extracted with three 15 mL quantities of water. The combined aqueous extracts were diluted to 50.0 mL and then shaken with 10 mL toluene. The toluene was then discarded. The absorbance of the aqueous solution when measured in a 1cm path length cell was 0.134 at 278 nm.Calculate the free chlorampenicol content of the sample in ppm, given A(1%, 1cm) = 298 at 278 nm.

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter17: Complexation And Precipitation Reactions And Titrations

Section: Chapter Questions

Problem 17.17QAP

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A 50.0 mL juice extract is colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the solution and the standard glucose solution (concentration: 1mg/mL) were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbance at 480 nm for the standard is 1.702 and for the sample, 0.926. A. What is the principle behind the assay?
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