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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 ?
in Chapter 2: Introduction to Analytical Chemistry by Expert (11.5k points) | 129 views

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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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