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\"Low concentrations of dioxin near the detection limit gave the following dimensionless instrument readings: 193.3, 156.5,...

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Chemistry

\"Low concentrations of dioxin near the detection limit gave thefollowing dimensionless instrument readings: 193.3, 156.5, 150.3,162.3, 207.3, 148.1, 211.1, 189.7, 122.7, and 193.5. Ten blanks hada mean reading of 47.3. The slope of the calibration curve is 2.55× 10x9 M–1. Estimate the signal and concentration detection limitsand the lower limit of quantitation for dioxin.\"

I know the signal detection limit is yblank + 3s and theconcentration limit is 3s/m, but what is ysample to calculatem?

Answer & Explanation Solved by verified expert
4.0 Ratings (720 Votes)

If 135.53 is standard deviation then,

Concentration detection limit = 3 x standard deviation/slope of line

                                              = 3 x 135.53/2.55 x 10^9

                                              = 1.59 x 10^-7 M

lower limit of quantification (LOQ) = 10 x standard deviation/slope of line

                                                      = 10 x 135.53/2.55 x 10^9

                                                       = 5.31 x 10^-7 M
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