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# quant - Chemistry 222 Fall 2007 Exam 1 Chapters 1-4 80...

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1 Chemistry 222 Name__________________________________________ Fall 2007 Exam 1: Chapters 1-4 80 Points Complete two (2) of problems 1-3 and four (4) of problems 4-8. CLEARLY mark the problem you do not want graded. You must show your work to receive credit for problems requiring math. Report your answers with the appropriate number of significant figures. Do two of problems 1-3. Clearly mark the problem you do not want graded. (10 pts each) 1. A Standard Reference Material is certified to contain 85.4 ppm of an organic contaminant in soil. You analyze this SRM to characterize a new method you are developing. Your analysis gives values of 88.6, 87.4, 83.6, 88.4, and 87.2 ppm. Do your results indicate the presence of systematic error in your method at the 95% confidence level? Justify your answer. With all of the other data bunched around 87 and 88 ppm, the point at 83.6 ppm should look a little odd and worthy of a Q-test. Q for 5 observations is 0.64 87.2-83.6 = 3.6 = 0.72 >0 .64 so omit 83.6 88.6-83.6 5 Based on the new dataset, the mean is 87.9 ppm, and s = 0.702 ppm To determine whether systematic error is indicated, determine if the “true value” falls within the confidence interval. (using the 95% confidence level). For 3 degrees of freedom and 95%, t table = 3.182 17 . 1 9 . 87 4 702 . 0 182 . 3 9 . 87 n ts 9 . 87 CI ± = × ± = ± = So, the confidence range is 88 ± 1 ppm, which does not include the true value, therefore, there seems to be an indication of systematic error (at least a 5% chance). If you do not do the Q-test, the confidence range becomes 87 ± 3 ppm 2. Concisely describe how you would prepare 1.00 L of a 100.0 ppm Pb 2+ solution from solid lead nitrate. What is the molar concentration of lead in this solution? You may assume a density of 1.00 g/mL for all solutions. There are multiple ways to approach this problem. Here’s one. Since d=1.00g/mL, we can approximate ppm with mg/L 100.0 mg Pb 2+ x 1 mol Pb 2+ x 1 g = 4.82 6 x10 -4 M Pb 2+ L solution 207.19 g Pb 2+ 1000 mg From this concentration, we can find the mass of Pb(NO 3 ) 2 needed. 4.82 6 x10 -4 mol Pb 2+ x 1 mol Pb(NO 3 ) 2 x 331.2098 g = 0.159 9 g Pb(NO 3 ) 2 L solution 1 mol Pb 2+ 1 mol Pb(NO 3 ) 2 1 L solution So, to prepare 1 L of 100.0 ppm Pb 2+ , dissolve 0.160 g of Pb(NO 3 ) 2 in water and dilute to 1.00L. This will make the lead concentration 4.83 x 10 -4 M.

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