Titration of 0.2342 g of pure Na,C,O, (134.0 g/mol) required 33.45 ml of KMNO, solution according to the chemical reaction: 2MNO, + 5C,0, +16 H* 2MN2+ + 10CO, +8H,0 The Molarity of KMNO, solution would be: a. 0.01609 M b. 0.04020 M c. 0.02090 M d. 0.20140 M e. 0.12450 M

Titration of 0.2342 g of pure Na,C,O, (134.0 g/mol) required 33.45 ml of KMNO, solution according to the chemical reaction: 2MNO, + 5C,0, +16 H* 2MN2+ + 10CO, +8H,0 The Molarity of KMNO, solution would be: a. 0.01609 M b. 0.04020 M c. 0.02090 M d. 0.20140 M e. 0.12450 M

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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Moles SCN- used in titration (mol)Moles of Agt in 20.0 mL titrated (mol) [Ag*lequil (M)X= [Ag*Jinit - [Ag*lequil = Change (M) solve for flask d e and f
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Molarity of titrant (NaOH): 0.4550 M HC2H3O2 (aq) + NaOH (aq) → NaC2H3O2 (aq) + H2O (l) Trial # First Second Third Fourth Initial buret reading 0.15 mL 2.43 mL 1.32 mL 0.58 mL Final buret reading 18.62 mL 20.87 mL 20.03 mL 19.14 mL Volume of titrant used 18.47 mL 18.44 mL 18.71 mL 18.56 mL 4) Calculate the molarity of the acetic acid in the vinegar solution (Show your work). use FW for moles-->grams acetic acid. Molarity acetic acid = _____________ M 5) Calculate the weight % of acetic acid in the vinegar. How does this compare with the % listed on the label (5.00%)? (For this calculation assume that density of vinegar is 1.03 g/mL and of course, show your work). Weight % = ___________ 6) If you didn’t get the same weight % of acetic acid as listed on the vinegar label (5.00 %), what are two things (be specific) that could’ve happened during the experiment that could explain the variation from the expected weight %? To do…