Internal standard. A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas of 3 473 and 10 222, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixed solution was diluted to 10.0 mL. This solution gave peak areas of 5 428 and 4 431 for X and S, respectively. (a) Find the response factor for X relative to S in Equation 5-9. (b) Find [S] (mM) in the 10.0 mL of mixed solution. (c) Find [X] (mM) in the 10.0 mL of mixed solution.(d) Find [X] in the original unknown.
Internal standard. A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas of 3 473 and 10 222, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixed solution was diluted to 10.0 mL. This solution gave peak areas of 5 428 and 4 431 for X and S, respectively. (a) Find the response factor for X relative to S in Equation 5-9. (b) Find [S] (mM) in the 10.0 mL of mixed solution. (c) Find [X] (mM) in the 10.0 mL of mixed solution.(d) Find [X] in the original unknown.
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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Internal standard. A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas of 3 473 and 10 222, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixed solution was diluted to 10.0 mL. This solution gave peak areas of 5 428 and 4 431 for X and S, respectively. (a) Find the response factor for X relative to S in Equation 5-9. (b) Find [S] (mM) in the 10.0 mL of mixed solution. (c) Find [X] (mM) in the 10.0 mL of mixed solution.
(d) Find [X] in the original unknown.
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