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How do I find the standard deviation of the molaR CONCENTRATION and the relative standard deviation? The values are in the picture

How do I find the standard deviation of the molaR CONCENTRATION and the relative standard deviation? The values are in the picture

Chemistry
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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How do I find the standard deviation of the molaR CONCENTRATION and the relative standard deviation? The values are in the picture

Trial 1 Trial 2 Trial 3 0.6163 g 0.623 g 0.6130 g 1. Tared mass of KHC3H,O4 (g) 2. Molar mass of KHC3H,O4 204.23 g/mol 3. Moles of KHC3H,O4 (mol) 0.6163 / 204.23 = 0.00302 0.623 / 204.23 = 0.00305 0.6130 / 204.23 = 0.00300 Titration apparatus approval 0.2 0.2 0.2 4. Buret reading of NaOH, initial (mL) 22.5 22.7 22.6 5. Buret reading of NaOH, final (mL) 6. Volume of NaOH dispensed (mL) 22.5 - 0.2 = 22.3 22.7 - 0.2 = 22.5 22.6 - 0.2 = 22.4 7. Molar concentration of NaOH (mol/L) 0.00302 /(22.3 * 10^-3) = 0.135 0.00305 /(22.5 * 10^-3) = 0.136 0.00300 /(22.4 * 10^-3) = 0.134 8. Average molar concentration of NaOH (mol/L) 0.135 Data Analysis, B 9. Standard deviation of molar concentration Data Analysis, C 10. Relative standard deviation of molar concentration (%RSD) Data Analysis, B
Transcribed Image Text:Trial 1 Trial 2 Trial 3 0.6163 g 0.623 g 0.6130 g 1. Tared mass of KHC3H,O4 (g) 2. Molar mass of KHC3H,O4 204.23 g/mol 3. Moles of KHC3H,O4 (mol) 0.6163 / 204.23 = 0.00302 0.623 / 204.23 = 0.00305 0.6130 / 204.23 = 0.00300 Titration apparatus approval 0.2 0.2 0.2 4. Buret reading of NaOH, initial (mL) 22.5 22.7 22.6 5. Buret reading of NaOH, final (mL) 6. Volume of NaOH dispensed (mL) 22.5 - 0.2 = 22.3 22.7 - 0.2 = 22.5 22.6 - 0.2 = 22.4 7. Molar concentration of NaOH (mol/L) 0.00302 /(22.3 * 10^-3) = 0.135 0.00305 /(22.5 * 10^-3) = 0.136 0.00300 /(22.4 * 10^-3) = 0.134 8. Average molar concentration of NaOH (mol/L) 0.135 Data Analysis, B 9. Standard deviation of molar concentration Data Analysis, C 10. Relative standard deviation of molar concentration (%RSD) Data Analysis, B
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