D a 14. The unknown tablet is prepared for analysis in two steps. The tablet is crushed into a powder and then dissolved into a 250 mL volumetric flask with dilute acetic acid. A diluted powder solution is prepared by transferring 25.00 mL of the original powder solution into a 50 mL volumetric flask and diluting to volume with dilute acetic acid. Then 1.00 mL of the diluted powder solution is added to each of the standard addition flasks; each flask has a total volume of 10.00 mL. The unknown's riboflavin concentration in the standard addition solutions is determined to be 1.41 ppm. What is the riboflavin concentration (ppm) in the unknown powder solution? 2.82 21.2 14.1 28.2

D a 14. The unknown tablet is prepared for analysis in two steps. The tablet is crushed into a powder and then dissolved into a 250 mL volumetric flask with dilute acetic acid. A diluted powder solution is prepared by transferring 25.00 mL of the original powder solution into a 50 mL volumetric flask and diluting to volume with dilute acetic acid. Then 1.00 mL of the diluted powder solution is added to each of the standard addition flasks; each flask has a total volume of 10.00 mL. The unknown's riboflavin concentration in the standard addition solutions is determined to be 1.41 ppm. What is the riboflavin concentration (ppm) in the unknown powder solution? 2.82 21.2 14.1 28.2

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter12: Solutions

Section: Chapter Questions

Problem 12.107QE

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Malic acid (C4H6O5) is a dicarboxylic acid that naturally occurs in the production of bignay wine. Maria was tasked to find out the acid content in a 200.0 mL wine sample by titrating it with a standardized solution of KOH. He was able to establish the working concentration of the KOH solution by using 39.12 mL of it to titrate a primary standard of KHP that weighed 3.021 g. The grams and %w/w of Malic acid present in the 200.0 mL wine sample were then determined by titrating a 20.00 mL aliquot using 48.95 mL of the standardized KOH solution. C4H6O5 (aq) + 2KOH (aq) ↔ K2C4H4O5 (aq) + 2H2O (l) Assume the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of C4H6O5 = 134.0874 g/mol What is the molar concentration of the prepared KOH solution?…
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