CHEM 4
DENSITY PROBLEMS
Show your work. Each number in your calculation should be accompanied by an appropriate unit. The
final answer should be recorded with the correct number of significant digits.
1. Use the handout “Densities of Some Common materials at 25ºC” to complete the table below.
Classified as:
Material
Element
carbon dioxide
Homogeneous
Mixture
X
X
air
X
olive oil
X
cement
Heterogeneous
Mixture
X
water
aluminum
Compound
X
2. Below is a graph showing the relationship between the mass and volume for water. Within the
same coordinate plane draw a graph showing the relationship between the mass and volume for:
(i) aluminum; (ii) ethyl alcohol. (Please use a ruler to draw the graphs.)
3. Below is a cube that approximately corresponds to one gram of water. Next to it draw
(approximately to the scale) two cubes: a cube of one gram of iron and a cube of one gram of
air.
4. Fill the blanks in the following.
1 mL H2O (at 4°C)
=
1.000
g H2O
1 cm3 H2O (at 4°C) =
1.000
g H2O
1L
H2O (at 4°C) =
1000
g H2O
1L
H2O (at 4°C) =
1.000
kg H2O
1000 L H2O (at 4°C) =
1000
kg H2O
1 m3
1000
kg H2O
H2O (at 4°C) =
5. Find the density of ethyl alcohol if 80.0 cm3 weighs 63.3 g.
Ans.: 0.791 g/cm3
Mass
Density =
63.3 g
=
Volume
80.0 cm
3
= 0.791 g/cm3
6. A piece of metal weighs 79.432 g. When the metal was placed into a graduated cylinder partially
filled with water the level of water went up from its original 35.5 mL to 46.5 mL. What is the density
of the metal in g/cm3?
Ans.: 7.22 g/cm3
Volume = 46.5 mL – 35.5 mL = 11.0 mL = 11.0 cm3
Mass
79.432 g
Density =
=
Volume
11.0 cm
= 7.22 g/cm3
3
7. Find the volume in liters of 40 kg of carbon tetrachloride, CCl4. (Density = 1.60 g/cm3.)
Ans.: 25 L
1 cm3
1000 g
40 kg
×
1 mL
×
×
1 kg
1 cm3
1.60 g
1L
×
= 25 L
1000 mL
8. The mass of fuel in an airplane must be carefully accounted for before a takeoff. If a 747 contains
155,211 L of fuel, what is the mass of the fuel in kilograms? Assume the density of the fuel to be
0.768 g/cm3? (N.J. Tro, page 51, problem 111).
Ans.: 119,000 kg
1000 cm3
155211 L
0.768 g
×
×
1L
1 cm
3
1 kg
×
= 119000 kg
1000 g
9. Osmium, element 76, is the densest natural element, with a density of 22.59 g/cm3 (twice that of
lead). What is its density in pounds per cubic inch?
Ans.: 0.8161 lb/in3
22.59 g
1 cm
3
(2.54 cm)3
1 lb
×
×
453.6 g
3
1 in
= 0.8161 lb/in3
10. An electrolytic tin-coating process gives the coating 30 millionths of an inch thick. How many
square meters can be coated with one kilogram of tin, density 7.30 g/cm3?
Ans.: 180 m2
2.54 cm
Thickness = 30×10-6 in ×
= 7.62×10-5 cm
1 in
Volume = Area × Thickness = Area × 7.62×10-5 cm
Volume =
Area =
Mass
Density
1000 g
7.30 g/cm3
=
Volume
Thickness
=
= 137 cm3
137 cm3
-5
7.62×10 cm
= 1.8×106 cm2
= 1.8×102 m2
11. An 56.1 cm3 sample of nickel weighs 0.500 kg. The density of lead is 11.3 g/cm3. How many
grams of nickel metal would occupy the same volume as 215 g of lead?
Ans.: 170. g or 1.70×102 g
Mass
0.500 kg
Density of Ni =
=
= 8.91 g/cm3
3
Volume
56.1 cm
Volume of Pb =
Mass of Ni =
Mass
Density
=
215 g
11.3 g/cm3
19.0 cm3 × 8.91 g/cm3
= 19.0 cm3
= 170. cm3
12. A quick method of determining density utilizes Archimedes' principle, which states that the
buoyant force on an immersed object is equal to the weight of the liquid displaced. A bar of
magnesium metal attached to a balance by a fine thread weighed 31.13 g in air and 19.35 g
when completely immersed in hexane (density 0.659 g/cm3). Calculate the density of this sample
of magnesium.
Ans.: 1.74 g/cm3
31.13 g – 19.35 g = 11.78 g = Archimedes’ Force = Weight of Hexane Displaced
Volume of Mg =
Volume of Hexane
Displaced
=
Density of Mg =
Mass of Mg
Volume of Mg
=
Mass of Hexane
Density of Hexane
11.78 g
17.9 cm3g
=
11.78 g
0.659 g/cm3
= 1.74 g/cm3
= 17.9 cm3