Question

At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements

Answer 1

Given that:

• At 25C the density of water is 0.997044 g/mL.

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.

• The change in the volume of the burette = (final volume - the initial volume) mL

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

$\mathbf{density = \dfrac{mass}{volume}}$

$\mathbf{density = \dfrac{5.01 g}{5.000 ml}}$

density = 0.998004 g/ml

Sample 2:

$\mathbf{density = \dfrac{5.025 g}{5.03ml}}$

density = 0.999006 g/ml

Sample 3:

$\mathbf{density = \dfrac{6.112 g}{6.13ml}}$

density = 0.997064 g/ml

Sample 4:

$\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}$

density = 0.999012 g/ml

Sample 5:

$\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}$

density = 0.997033 g/ml

Thus, the average density for all the samples is:

$\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 + 0.999012 + 0.997033 )}{5}}$

= 0.998024

∴

The percentage error for the two densities measurement is:

$=\dfrac{ (experimental \ value -theoretical \ value)\times 100 }{theoretical \ value}$

Given that the theoretical value = 0.997044 g/ml

Then;

$\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}$

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

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