*) mass = 9.9876 (+or-) 0.0004 g **) mass = 9.9866 (+or-) 0.0004 g ***) mass = 9

Question

*) mass = 9.9876 (+or-) 0.0004 g

**) mass = 9.9866 (+or-) 0.0004 g

***) mass = 9.9873 (+or-) 0.0004 g

****) mass = 9.9878 (+or-) 0.0004 g

*****) mass = 9.9864 (+or-) 0.0004 g

A student was calibrating a 10-mL pipette which was guaranteed by manufacturers to deliver 10.00(+or-) 0.02 mL if used correctly. He measured the mass of water delivered by the pipette five times and obtained the masses shown on top. the mean mass is 9.98714 g.

1)what is the absolute uncertainty associated with this mean mass? (use 1/2 range)

2) he used a conversion factor (1g = 1.0021 mL) to convert mass to volume. what is the mean value?

3) how do you decide?

4) after you decide, calculate the mean volume and its absolute uncertainty. show your working, include all units and round off appropriately.

Explanation / Answer

1) Uncertainty associated with the individual mass measurements = 0.0004 g; mean mass of the water pipetted = 9.98714 g.

Given the individual measurements and the mean measurement, find the variance = [(9.9876 – 9.98714)2 + (9.9866 – 9.98714)2 + (9.9873 – 9.98714)2 + (9.9878 – 9.98714)2 + (9.9864 – 9.98714)2] = 1.512*10-6

Find the standard deviation = (1.512*10-6)/5 = 5.499*10-4 = 0.0005499 0.0005

The absolute error in the measurement of the mean is 0.0005 g (ans).

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.