When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 10.055 g at 20ºC. If the density of water is taken to be 0.9975 g/mL at 20ºC, what is the percent error for the 10 mL of water.

Answer is: percent error for the 10 mL of water is 0,795 %.
d(water) = 0,9975 g/mL.
V(water) = 10 mL.
m(water) = d(water) · V(water).
m(water) = 0,9975 g/mL · 10 mL.
m(water) = 9,975 g.
error = 10,055 g - 9,975 g ÷ 10,055 g · 100%
error = 0,795%.
d - density.
V - volume.

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maacastrobr maacastrobr · Answer 2 · 2019-10-04T22:47:52+02:00

maacastrobr maacastrobr

04-10-2019

Answer:

%error = 0.802%

Explanation:

We use the formula

%error =[tex]\frac{|AcceptedValue-ExperimentalValue|}{AcceptedValue}[/tex] * 100

Where the accepted value is 10 mL, and the experimental value is calculated using the formula for density:

density=mass/volume (ρ=m/V)

Experimental Volume = 10.055 g / 0.9975 g·mL⁻¹ = 10.080 mL

%error= [tex]\frac{|10 -10.080|}{10}[/tex]*100%

%error = 0.802%

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