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A solution containing 3.39 mM X (analyte) and 1.72 mM S (internal standard) gave peak areas of 3497 and 9997, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 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 5428 and 4431 for X and S, respectively.
a. Calculate the response factor for the analyte. (keep four significant figures)
b. Find the concentration (mM) of X in the original unknown
c. Find the concentration (mM) of X in the 10.0 mL of mixed solution.

Sagot :

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Answer:

a. Response factor is 0.1775

b. 11.7mM of X in the original unknown

c. 5.85mM is the concentration of X in the diluted solution

Explanation:

a. The equation to find response factor is:

F = AreaX*[S] / AreaS*[X]

Where Area of X our sample and S the internal standard whereas [] represent the concentration of each one.

Replacing:

F = 3497*1.72mM / 9997*3.39mM

F = 0.1775

c. The concentration of S is:

8.47mM * 1.00mL / 10.0mL = 0.847mM

Replacing:

0.1775 = 5428*0.847mM / 4431*[X]

[X] = 5.85mM is the concentration of X in the diluted solution

b. The concentration of X in the original unknown is:

5.85mM * (10.0mL / 5.00mL) =

11.7mM of X in the original unknown

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