124) The method of internal standard was used toanalyze and measure the concentration of chemicalX by GCA known mixture of compounds X and Y (internalstandard) gave the following chromatographicresults:CompoundXYConcentration(M)6.3 x 10-82.0 x 10-7a) 2.0 x 10-6 Mb) 8.4 x 10-7 Mc) 6.6 x 10-7 Md) 4.1 x 10-7 Me) 2.9 x 10-7 MPeak Area395787Then, an unknown was analyzed as follows.3.0 mL of unknown X was treated with 0.100 mLof 1.6 x 10-5 M Y and the mixture diluted to 10.0mL final volume. Peak areas were 633 (for X) and520 (for Y). What is the concentration of X in theunknown. (The original 3.00 sample?)

Volume of unknown X =3.0 mL Concentration of Y in the mixture = 1.6 x 10-5 M Volume of Y in the…

Answered: 24) The method of internal standard was…

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

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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Question

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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