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प्रश्न
Solve the following problem:
A 1.000 mL sample of acetone, a common solvent used as a paint remover, was placed in a small bottle whose mass was known to be 38.0015 g.
The following values were obtained when the acetone - filled bottle was weighed: 38.7798 g, 38.7795 g and 38.7801 g. How would you characterise the precision and accuracy of these measurements if the actual mass of the acetone was 0.7791 g?
उत्तर
i. Precision:
|
Measurement |
Mass of acetone observed (g) |
| 1 | 38.7798 – 38.0015 = 0.7783 |
| 2 | 38.7795 – 38.0015 = 0.7780 |
| 3 | 38.7801 – 38.0015 = 0.7786 |
Mean = `(0.7783+0.7780+0.7786)/3` = 0.7783 g
|
Measurement |
Mass of acetone observed (g) | Absolute deviation (g) = |Observed value – Mean| |
| 1 | 0.7783 | 0 |
| 2 | 0.7780 | 0.0003 |
| 3 | 0.7786 | 0.0003 |
Mean absolute deviation = `(0+0.0003+0.0003)/2` = 0.0002
∴ Mean absolute deviation = ±0.0002 g
Relative deviation = `"Mean absolute deviation"/ "Mean"xx100%`
= `0.0002/0.7783xx100%`
= 0.0256%
ii. Accuracy:
Actual mass of acetone = 0.7791 g
Observed value (average) = 0.7783 g
a. Absolute error = Observed value – True value
= 0.7783 – 0.7791
= – 0.0008 g
b. Relative error = `"Absolute error"/"True value"xx 100%`
= `(-0.0008)/0.7791xx100%`
= –0.1026%
These observed values are close to each other and are also close to the actual mass. Therefore, the results are precise and as well accurate.
- Relative deviation = 0.0256%
- Relative error = 0.1026%
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