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

The percent error for the 10 mL of water, given a measured mass of 9.925 g and a density of 0.9975 g/mL, is approximately 0.501%.

Step-by-step explanation:

To calculate the percent error for the 10 mL of water given the measured mass and the density of water, we can use the formula for percent error:

Percent Error = | (Experimental Value - Theoretical Value) / Theoretical Value | × 100%

The theoretical mass of 10 mL of water using the density 0.9975 g/mL is calculated as follows:

Theoretical Mass = Volume × Density = 10 mL × 0.9975 g/mL = 9.975 g

The experimental mass measured is 9.925 g. Now, applying the percent error formula:

Percent Error = | (9.925 g - 9.975 g) / 9.975 g | × 100% = |(-0.05 g) / 9.975 g| × 100% ≈ 0.501%

Therefore, the percent error for the measurement of 10 mL of water is approximately 0.501%.

Answer 2

measured volume = 10 ml
mass = 9.925 g
density = 0.9975 g/ml
density =
`(mass)/(volume)`
so actual volume =
`(mass)/(density)` =
`(9.925)/(0.9975)` = 9.95 mL
Percentage Error =
`(measured volume - Actual volume)/(actual volume)` x 100
=
`(10 - 9.95)/(9.95)` x 100 = 0.5 % error