Table 2.2, The density of solids.
Aluminum (bar) 39 length = 9.7 width = 1.6 height = 1.0 (vol) 40 - 25 = 15 (density)
PE (rod) 21 diameter = 1.7 length = 10.5 (vol) 48 - 25 = 23 (density)
Iron(bolt) 39 XXX 31.9 - 25 = 6.9 XXX
Note: Vol. = volume. Since the bolt has an irregular shape, find its volume by displacement only.
1. Calculate the volumes of the samples used in Procedure step 2. Record the calculated
volumes in Table 2.2. Why could we not find the volume of the bolt by this method?
2. How did you find the volume of the PE rod using the displacement method?
Note: 1 cm3 = 1 mL.
Because PE floats, push the rod under the water with a very small object, such as a
pin, and volume of the pin is negligible and can be ignored.
3. Except for the bolt, note the difference between the calculated volumes and the volume
found by the displacement method. Which method is more accurate? Why?
Mass, Volume, and Density
4. Use the volumes found by displacement to calculate the densities of the objects.
Record your data in Table 2.2.
5. The accepted values for density are: aluminum = 2.71 g/cm3; PE = about 0.925 g/cm3; and iron = 7.87 g/cm3. Compare your values from Question 4 with these. How do they compare? Why are they different? Note: the density of different kinds of PE and PVC vary because of the difference in the arrangements of their molecules.
Table 2.3, The density of liquids
Liquid Volume, mL Mass, g Density, g/mL
Water 50 77 - 27 = 50
Saturated Salt Solution 50 83 - 27 = 56
6. Calculate the densities of the liquids in Table 2.3. The accepted density of water is 1 g/
cm3. How do your results compare? Remember that 1 cm3 = 1 mL.
7. How does the density of water compare with the density of the saturated salt solution? Is this what you would have predicted?
8. Length of room __4.02__ m
9. Width of room __3.37__ m
10. Height of room __2.35__ m
11. Volume of room ________ m3
12. Mass of air in room _______ kg
13. Are you surprised by the mass of air in the room? Explain.
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