The three identical samples of potassium chlorate are decomposed. The mass of oxygen is determined to be 3.87 g, 3.95 g and 3.89 g for the set.
The three identical samples of potassium chlorate are decomposed. The mass of oxygen is determined to be 3.87 g, 3.95 g and 3.89 g for the set. Calculate absolute deviation and relative deviation.
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Mean = \(\frac{3.87+3.95+3.89}{3}\)
= 3.90
| Sample | Mass of oxygen | Absolute deviation = | Observed value – Mean | |
| 1 | 3.87 g | 0.03 g |
| 2 | 3.95 g | 0.05 g |
| 3 | 3.89 g | 0.01 g |
Mean absolute deviation = \(\frac{0.03+0.05+0.01}{3}\)
= 0.03
∴ Mean absolute deviation = \(\pm0.03g\)
Relative deviation = \(\frac{Mean\,absolute\,deviation}{Mean}\) x 100%
= \(\frac{0.03}{3.90}\) x 100%
= 0.8%
∴ i. Absolute deviation in each observation = 0.03 g, 0.05 g, 0.01 g
Relative deviation = 0.8%
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