A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H2C to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the tablet? KIO3 + 5 KI + 6 H* = 3 12 + 6 K* + 3 H20 C6H8O6 + 12 = C6H606 + 21 + 2H* 3.39 % 33.89 % 67.77 %
A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H2C to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the tablet? KIO3 + 5 KI + 6 H* = 3 12 + 6 K* + 3 H20 C6H8O6 + 12 = C6H606 + 21 + 2H* 3.39 % 33.89 % 67.77 %
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H20
to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was
analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the
tablet?
KIO3 + 5 KI + 6 H* = 3 12 + 6 K* + 3 H20
C6H8O6 + 12 = C6H6O6 + 21 + 2H*
%3D
3.39 %
33.89 %
67.77 %
22.59 %
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