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A solution was prepared by mixing 1.00 mL of unknown (X) with 5.00mL of standard (S)...

A solution was prepared by mixing 1.00 mL of unknown (X) with 5.00mL of standard (S) containing 8.24 μg S/mL and diluting to 50.0 mL. The measured signal quotient (signal due to X/signal due to S) was 1.69. In a separate experiment, it was found that, for equal concentrations of X and S, the signal due to X was 0.930 times as intense as the signal due to S. Find the concentration of X in the unknown

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Answer #1

signal X is 0.93 times signal S when concentrations are same

Let X conc be [X] and S conc be [S]

given signal of X / signal of S = 1.69

hence 0.93 [X] /[S] = 1.69

[X]/[S] = 1.817

S mass =8.24 ug/ml x 5 ml = 41.2 ug

[S] = 41.2 ug /50 ml = 0.824 ug /ml

hence by ratio [X]/ 0.824 = 1.817

[X] = 1.4974 = ug of X / 50 ml

ug of X ( mass of X in ug) = 74.87            alll this X is due to 1ml added

Hence [X] = 74.87 ug of X /ml

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