A solution containing 2.50 mM X (analyte) and 1.20 mM S (standard) gave peak areas of 2509 and 7132 respectively, in a chromatographic analysis. Then, 1.00 mL of 6.00 mM S was added to 5.00 mL of unknown X and the mixture was diluted to 10.0 mL. This solution gave peak areas of 4016 and 3138 for X and S, respectively. Calculate i. the response factor for the analyte ii. the concentration of S (mM) in the 10.0 mL of mixed solution iii. the concentration of X (mM) in the 10.0 mL of mixed solution
A solution containing 2.50 mM X (analyte) and 1.20 mM S (standard) gave peak areas of 2509 and 7132 respectively, in a chromatographic analysis. Then, 1.00 mL of 6.00 mM S was added to 5.00 mL of unknown X and the mixture was diluted to 10.0 mL. This solution gave peak areas of 4016 and 3138 for X and S, respectively. Calculate i. the response factor for the analyte ii. the concentration of S (mM) in the 10.0 mL of mixed solution iii. the concentration of X (mM) in the 10.0 mL of mixed solution
Chapter88: Column Chromatography
Section: Chapter Questions
Problem 5P
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3c.
A solution containing 2.50 mM X (analyte) and 1.20 mM S (standard) gave peak areas of 2509 and 7132 respectively, in a chromatographic analysis. Then, 1.00 mL of 6.00 mM S was added to 5.00 mL of unknown X and the mixture was diluted to 10.0 mL. This solution gave peak areas of 4016 and 3138 for X and S, respectively.
Calculate
i. the response factor for the analyte
ii. the concentration of S (mM) in the 10.0 mL of mixed solution
iii. the concentration of X (mM) in the 10.0 mL of mixed solution
iv. the concentration of X in the original unknown solution
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