Homework Help is Here – Start Your Trial Now!
Science
Chemistry
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...
See similar textbooks
icon
Related questions
Question

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
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Statistics and Analytical Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS