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.

Question

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 ?

Answer 1

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%