Question

When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 9.925 g. If the density of water is taken to be 0.9975 g/mL, what is the percent error for the 10 mL of water

Answer 1

Are you having trouble with a particular part of this problem? We need to find the density of water from our mass and volume measured, and then compare it to the actual assumed density given 0.9975 g/mL.

Answer 2

i can't figure out the whole q

Answer 3

The percent error is given as:

\( \displaystyle \rm \% \ error = \frac{|Actual - Measured|}{Actual} \times 100\% \)
We know the actual density because it is given: 0.9975 g/mL.
The density in terms of mass and volume is: Density = Mass / Volume.
We have a mass: 9.925 g
We have a volume: 10 mL
So the density is: Density = 9.925 g / 10 mL which we can simplify.

Then it all goes into the % error formula and out pops a percent error:
\( \displaystyle \rm \% \ error = \frac{|0.9975 - 0.9925|}{0.9975} \times 100\% \)

Answer 4

thank you so much

Answer 5

No problem! :)