3. Seven analyses for the phosphorus content of a fertilizer resulted in 16.2, 17.5, 15.4, 15.9, 16., 16.3, and 17.1%. Find (a) the standard deviation, S, and (b) the 95% confidence interval for the true value, H. 4. The following five values were obtained for the wt % of an organic acid in a sample: 30.3, 31.1,

Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
ChapterA1: Evaluation Of Analytical Data
Section: Chapter Questions
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3. Seven analyses for the phosphorus content of a fertilizer resulted in 16.2, 17.5, 15.4, 15.9, 16.8,
16.3, and 17.1%. Find (a) the standard deviation, s, , and (b) the 95% confidence interval for
the true value, H.
4. The following five values were obtained for the wt % of an organic acid in a sample: 30.3, 31.1,
32.6, 36.7, 28.9. Determine if the value of 36.7 may be rejected at the 90% (two-tailed) confidence
level.
5. The concentration of a NaOH solution is determined in a single experiment by titrating a weighed
sample of pure, dried KHP (MW = 204.229) with the NaOH. The data are:
Weight of KHP = (11.6723 + 0.0001) - (10.8364 ±0.0001) g
Volume NaOH = (32.68 + 0.02) - (1.24 + 0.02) mL
Where the standard deviations are o, values. (a) What is the molarity of the NaOH?
(b) Based on the propagation-of-error formula, what is the standard deviation of the molarity,oM ?
Transcribed Image Text:3. Seven analyses for the phosphorus content of a fertilizer resulted in 16.2, 17.5, 15.4, 15.9, 16.8, 16.3, and 17.1%. Find (a) the standard deviation, s, , and (b) the 95% confidence interval for the true value, H. 4. The following five values were obtained for the wt % of an organic acid in a sample: 30.3, 31.1, 32.6, 36.7, 28.9. Determine if the value of 36.7 may be rejected at the 90% (two-tailed) confidence level. 5. The concentration of a NaOH solution is determined in a single experiment by titrating a weighed sample of pure, dried KHP (MW = 204.229) with the NaOH. The data are: Weight of KHP = (11.6723 + 0.0001) - (10.8364 ±0.0001) g Volume NaOH = (32.68 + 0.02) - (1.24 + 0.02) mL Where the standard deviations are o, values. (a) What is the molarity of the NaOH? (b) Based on the propagation-of-error formula, what is the standard deviation of the molarity,oM ?
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