The following data were obtained for a voltage measurement of the same sample on a noisy system:| Measurement # | Value (mV) ||---------------|------------|| 1 | 1.2 || 2 | 1.08 || 3 | 1.16 || 4 | 1.15 || 5 | 1.18 || 6 | 1.11 || 7 | 1.08 || 8 | 1.25 |The average is 1.15125 and the standard deviation is 0.059626815635728.If the slope of the calibration curve is 52.02 mV/ppm, what is the concentration detection limit?\[ \text{Detection Limit (D.L.)} = \frac{3s}{\text{slope}} \]\[ \text{Lower Limit of Quantification (LLoQ)} = \frac{10s}{\text{slope}} \]\[ \text{Concentration Detection Limit} = \underline{\hspace{3cm}} \text{ ppm (answer using 3 significant digits)} \] The following data were obtained for a voltage measurement of the same sample on a noisy system:| Measurement # | Value (mV) ||---------------|------------|| 1 | 1.20 || 2 | 1.08 || 3 | 1.16 || 4 | 1.15 || 5 | 1.18 || 6 | 1.11 || 7 | 1.08 || 8 | 1.25 |The average is 1.15125 and the standard deviation is 0.059626815635728.If the slope of the calibration curve is 52.02 mV/ppm, what is the lower limit of quantitation?\[ \text{LLoQ} = \frac{10s}{\text{slope}}\]Enter the answer in ppm (using 3 significant digits).

Answered: The average is 1.15125 and the standard…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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The following data were obtained for a voltage measurement of the same sample on a noisy system:

| Measurement # | Value (mV) |
|---------------|------------|
| 1 | 1.2 |
| 2 | 1.08 |
| 3 | 1.16 |
| 4 | 1.15 |
| 5 | 1.18 |
| 6 | 1.11 |
| 7 | 1.08 |
| 8 | 1.25 |

The average is 1.15125 and the standard deviation is 0.059626815635728.

If the slope of the calibration curve is 52.02 mV/ppm, what is the concentration detection limit?

\[ \text{Detection Limit (D.L.)} = \frac{3s}{\text{slope}} \]

\[ \text{Lower Limit of Quantification (LLoQ)} = \frac{10s}{\text{slope}} \]

\[ \text{Concentration Detection Limit} = \underline{\hspace{3cm}} \text{ ppm (answer using 3 significant digits)} \]

The following data were obtained for a voltage measurement of the same sample on a noisy system:

| Measurement # | Value (mV) |
|---------------|------------|
| 1 | 1.20 |
| 2 | 1.08 |
| 3 | 1.16 |
| 4 | 1.15 |
| 5 | 1.18 |
| 6 | 1.11 |
| 7 | 1.08 |
| 8 | 1.25 |

The average is 1.15125 and the standard deviation is 0.059626815635728.

If the slope of the calibration curve is 52.02 mV/ppm, what is the lower limit of quantitation?

\[
\text{LLoQ} = \frac{10s}{\text{slope}}
\]

Enter the answer in ppm (using 3 significant digits).

Expert Solution

Concentration of detection limit (D.L)

1) Concentration of detection limit (D.L)/Limit of detection (LOD): It is the smallest concentration that is formed when measured signal like voltage differs significantly from the background and can be easily distinguished from zero.

Standard deviation of the curve 1, s is 0.059626815635728 mV

Slope of the calibration curve, m = 52.02 mV/ppm

Concentration of detection limit, D.L = 3s/slope = 3 x standard deviation/slope

D.L = 3 x 0.059626815635728 mV/52.02 mV/ppm = 0.1788804469071 ppm/52.02 = 0.00343860228 ppm = 0.00344 ppm (3 significant figures)

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