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24) The method of internal standard was used to analyze and measure the concentration of chemical X by GC A known mixture of compounds X and Y (internal tandard) gave the following chromatographic

24) The method of internal standard was used to analyze and measure the concentration of chemical X by GC A known mixture of compounds X and Y (internal tandard) gave the following chromatographic

Chemistry
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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1 24) The method of internal standard was used to analyze and measure the concentration of chemical X by GC A known mixture of compounds X and Y (internal standard) gave the following chromatographic results: Compound X Y Concentration (M) 6.3 x 10-8 2.0 x 10-7 a) 2.0 x 10-6 M b) 8.4 x 10-7 M c) 6.6 x 10-7 M d) 4.1 x 10-7 M e) 2.9 x 10-7 M Peak Area 395 787 Then, an unknown was analyzed as follows. 3.0 mL of unknown X was treated with 0.100 mL of 1.6 x 10-5 M Y and the mixture diluted to 10.0 mL final volume. Peak areas were 633 (for X) and 520 (for Y). What is the concentration of X in the unknown. (The original 3.00 sample?)
Transcribed Image Text:1 24) The method of internal standard was used to analyze and measure the concentration of chemical X by GC A known mixture of compounds X and Y (internal standard) gave the following chromatographic results: Compound X Y Concentration (M) 6.3 x 10-8 2.0 x 10-7 a) 2.0 x 10-6 M b) 8.4 x 10-7 M c) 6.6 x 10-7 M d) 4.1 x 10-7 M e) 2.9 x 10-7 M Peak Area 395 787 Then, an unknown was analyzed as follows. 3.0 mL of unknown X was treated with 0.100 mL of 1.6 x 10-5 M Y and the mixture diluted to 10.0 mL final volume. Peak areas were 633 (for X) and 520 (for Y). What is the concentration of X in the unknown. (The original 3.00 sample?)
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