I know I\'m making this harder than it is, but I can\'t seem to get this. Consid

Question

I know I'm making this harder than it is, but I can't seem to get this.

Consider a solution containing 3.54 mlvl of an analysis , X, and 1.89 mM of a standard, S. Upon chromatographic separation of the solution peak areas for X and S are 3093 and 10515, repectively. Determine the response factor for X relative to S. To determine the concentration of X in an unknown slution, 1,00 mL of 8.49 mM S was added to 4.CO mL of the unknown X solution and the mixture was diluted to 10.0 mL. After chromatographic separation, this solution gave peak areas of 6239 and 4691 for X and S, respectively. Determine the concentration of S in the 10.0 mL solution.

Explanation / Answer

a)

rf = 3093 / 10515 = 0.29

b)

concentration of s is

M1V1 = M2 V2 M1 is 8.49 and V1 is 1 and V2 is 10

0.849

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