estion 20 of 20 Consider a solution containing 4.52 mM of an analyte, X, and 1.8

Question

estion 20 of 20 Consider a solution containing 4.52 mM of an analyte, X, and 1.87 mM of a standard, S. Upon chromatographic separation of the solution peak areas for X and S are 3517 and 10127, respectively. Determine the response factor for X relative to S Map dab Number To determine the concentration of X in an unknown solution, 1.00 mL of 8.52 mM S was added to 3.00 mL of the unknown X solution and the mixture was diluted to 10.0 mL. After chromatographic separation, this solution gave peak areas of 5313 and 4789 for X and S, respectively Determine the concentration of S in the 10.0 mL solution. Number mM Determine the concentration of X in the 10.0 mL solution Number mM Determine the concentration of X in the unknown solution Number Hint Check Answer Next el Exit

Explanation / Answer

a)

Peak area of X, Ax = 3517

Peak area of S, As = 10127

[X] = 4.52 mM

[S] = 1.87 mM

Response factor, F = Ax / [X] / (As / [S])

= 3517 / 4.52 / (10127 / 1.87)

F = 0.14

b)

Let final concentration of S be [S]f

Vf [S]f = V1 * M1

10 mL * [S]f = 1 mL * 8.52

[S]f = 0.852 mM

Ax = 5313

As = 4789

F = 0.14 = Ax / [X]f / (As / [S]f)

[X]f = Ax / (0.14 * As / [S]f)

[X]f = 6.58 mM

In unknown solution,

M1 V1 = M2 V2

M1 * 3 mL = 6.58 mM * 10 mL

M1 = 21.9 mM

[X] in unknown solution, M1 = 21.9 mM

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