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4 ) Solve the following problems

S. 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 ?
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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=$$\frac{0.7783 + 0.7780+ 0.7786}{ 3}=0.7783$$ 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 =$$\frac{0+0.0003+0.0003}{3}=0.0002$$

Mean absolute deviation =$$\pm0.0002 g$$

Relative deviation = $$\frac{Mass \space Absolute \space deviation}{ Mean}\times 100 \%$$

=$$\frac{0.0002}{0.7783}\times 100\% =0.0257\%$$

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 =$${Absolute \space error\over True \space Value }\times 100\% =-\frac{ 0.0008}{0.7791}\times100\%$$ =-0.1027%

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