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Given that:
From the information attached below, we have the following parameters.
The density of water calculation using a bottle.
Initial volume of Final volume of Mass of water Density (g/mL)
burette (mL) burette (mL) dispensed (g)
Sample 1 2.33 7.34 5.000 -----
Sample 2 7.34 12.37 5.025 -----
Sample 3 12.37 18.50 6.112 -----
Sample 4 18.50 24.57 6.064 -----
Sample 5 24.57 31.31 6.720 -----
The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.
Sample 1:
= (7.34 - 2.33) mL
= 5.01 mL
Sample 2:
= (12.37 - 7.34) mL
= 5.03 mL
Sample 3:
= (18.50 - 12.37) mL
= 6.03 mL
Sample 4:
= (24.57 - 18.50) mL
= 6.07 mL
Sample 5:
= (31.31 - 24.57) mL
= 6.74 mL
The mass of the water dispersed in sample 1 is given as = 5.000 g
Using the relation for calculating the density of each, we have:
Sample 1
[tex]\mathbf{density = \dfrac{mass}{volume}}[/tex]
[tex]\mathbf{density = \dfrac{5.01 g}{5.000 ml}}[/tex]
density = 0.998004 g/ml
Sample 2:
[tex]\mathbf{density = \dfrac{5.025 g}{5.03ml}}[/tex]
density = 0.999006 g/ml
Sample 3:
[tex]\mathbf{density = \dfrac{6.112 g}{6.13ml}}[/tex]
density = 0.997064 g/ml
Sample 4:
[tex]\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}[/tex]
density = 0.999012 g/ml
Sample 5:
[tex]\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}[/tex]
density = 0.997033 g/ml
Thus, the average density for all the samples is:
[tex]\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 + 0.999012 + 0.997033 )}{5}}[/tex]
= 0.998024
∴
The percentage error for the two densities measurement is:
[tex]=\dfrac{ (experimental \ value -theoretical \ value)\times 100 }{theoretical \ value}[/tex]
Given that the theoretical value = 0.997044 g/ml
Then;
[tex]\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}[/tex]
= 0.0983%
Therefore, we can conclude that the percent error for the two density measurements is 0.0983%
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