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A solution containing 2.54 mM X (analyte) and 1.24 mM S (standard) gave peak areas of...

A solution containing 2.54 mM X (analyte) and 1.24 mM S (standard) gave peak areas of 3473 and 1022, respectively, in a chromatographic analysis. Then 1.00 mL of 4.78 mM S was added to 2.50 mL of unknown X, and the mixture was diluted to 10.0 mL. This solution gave peak areas of 4864 and 4431 for X and S, respectively . Find the concentration of X (mM) in the 10.0 mL of mixed solution, and find the concentration of X in the original unknown.

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Answer #1

Area of analyte signal/concetration of analyte=F*(area of standard signal/concentration of standard signal)

From the first set of values we can clear F

3473/2.54=1022/1.24*F---> F=1.659

Then, knowing F we can use the same formula to calculate the concentration of X we the second set of values

The concentration of the Standartd solution (S) in the secon case is 0.001L*4.78mM/0.010L=0.478mM

4864/0.478=(1.659)*(4431/X)

Clearing X---> X=0.729mM

Concentration of X in the unknow solution is 0.010*0.729/0.0025=2.916mM

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