1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer statement? Show your work.

General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter4: Chemical Reactions
Section4.8: Diluting Solutions
Problem 4.6CC: Consider the following beakers. Each contains a solution of the hypothetical atom X. a Arrange the...
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Exercise 1
Data Table 1: NaOH Titration Volume
Trial 1
Trial 2
Trial 3
Initial NaOH
Volume (mL)
8.78
9.25
9.30
Final NaOH
Volume (mL)
????
Questions
0.30
0.98
1.01
Average of
Trials 1-3:
Total Volume of
NaOH Used (mL)
8.48
Average Volume of Concentration CH₂COOH
NaOH Used (mL) in vinegar (mol/L)
0.835
8.35
8.27
8.29
Data Table 2: Concentration of CH₂COOH in Vinegar
8.35
% CH₂COOH
in vinegar
4.76%
Type your calculations of concentration and percent vinegar here (no photos):
8.48 +8.27+8.29=25.04/3=8.346 → 8.35 mL
8.35/10=0.835 mol/L
1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0%
acetic acid. What is the percent error between your result and the manufacturer statement? Show your
work.
Transcribed Image Text:Exercise 1 Data Table 1: NaOH Titration Volume Trial 1 Trial 2 Trial 3 Initial NaOH Volume (mL) 8.78 9.25 9.30 Final NaOH Volume (mL) ???? Questions 0.30 0.98 1.01 Average of Trials 1-3: Total Volume of NaOH Used (mL) 8.48 Average Volume of Concentration CH₂COOH NaOH Used (mL) in vinegar (mol/L) 0.835 8.35 8.27 8.29 Data Table 2: Concentration of CH₂COOH in Vinegar 8.35 % CH₂COOH in vinegar 4.76% Type your calculations of concentration and percent vinegar here (no photos): 8.48 +8.27+8.29=25.04/3=8.346 → 8.35 mL 8.35/10=0.835 mol/L 1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer statement? Show your work.
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