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APPENDIX E

Solutions for Group A Parallel Exercises

600 g
Note: Solutions that involve calculations of measurements are _

V =
g
_
rounded up or down to conform to the rules for significant fig- 3.00

ures described in appendix A. cm 3
600 _
_ g _ 3
cm
= ×

3.00 1 g
CHAPTER 1 . 3
g cm
_

= 200
1.1. Answers will vary but should have the relationship of 100 cm in g
1 m, for example, 178 cm = 1.78 m. 3
= 200 cm
1.2. Since density is given by the relationship ρ = m/V, then
3
1.5. A 50.0 cm sample with a mass of 34.0 grams has a density of
272 g
m _ _

ρ = = m _

V 20.0 cm 3 ρ =
V
272 _
_ g 34.0 g

= _

20.0 cm 3 =
50.0 cm 3
g
_
= 13.6 _ g
34.0 _

cm 3 =
50.0 cm 3
1.3. The volume of a sample of lead is given and the problem asks _

g

for the mass. From the relationship of ρ = m/V, solving for the = 0.680
3
cm
mass (m) gives you the density (ρ) times the volume (V), or
3
3
m = ρV. The density of lead, 11.4 g/cm , can be obtained from According to Table 1.3, 0.680 g/cm is the density of gasoline,

Table 1.3, so so the substance must be gasoline.
m _ 1.6. The problem asks for a mass and gives a volume, so you need a

ρ =
V relationship between mass and volume. Table 1.3 gives the
3
mV
_ density of water as 1.00 g/cm , which is a density that is easily
Vρ =

V remembered. The volume is given in liters (L), which should
3

m = ρV first be converted to cm because this is the unit in which
density is expressed. The relationship of ρ = m/V solved for

g
3)
_
(
3
m = 11.4 (10.0 cm ) mass is ρV, so the solution is
cm
m _
g
_ 3 ρ = ∴ m = ρV

= 11.4 × 10.0 × cm V

cm 3
g
(
_
3)
g·cm
3
_ 3 m = 1.00 (40,000 cm )

= 114 cm
cm 3
g
_
= 114 g = 1.00 × 40,000 × cm 3

cm 3
1.4. Solving the relationship ρ = m/V for volume gives g·cm 3

V = m/ρ, and = 40,000 _

cm 3
m _
ρ = = 40,000 g
V
= 40 kg
_
mV
Vρ =

V
Vρ
_ m _

ρ
ρ =
m _

V =
ρ
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