Practice Problems:
Conversions and Dimensional Analysis
KEY
CHEM 1A
Conversion Factors
Length
Mass
Volume
1 m = 1.0936 yds
1 in = 2.54 cm (exactly)
1 km = 0.62137 mi
1 mi ≡ 5280 ft
1 kg = 2.205 lbs
1 lb = 453.59 g
1 ton ≡ 2000 lbs
1 metric ton (tonne) ≡ 1000 kg
1 L = 1.0567 qt
1 gal = 3.785 L
1 fl oz = 29.6 mL
1 cm3 ≡ 1 mL
Part I. Use dimensional analysis and one continuous string of conversion factors to solve the
following problems. Be sure to use complete units throughout.
1. How many micrograms (g) are in 9.17 kilograms (kg)?
103 g
1 g
1 kg
–6
10 g
∞ sig. fig.s
∞ sig. fig.s
9.17 kg
3 sig. fig.s
= 9.17 x 109 g
Note: Labeling each number and conversion
factor with significant figures is not part
of the required work, but was done to
illustrate how the significant figures of
each answer were obtained.
9.17 x 109 g
Answer: _______________________
2. How many cubic centimeters (cm3) are in 2.5 gallons (gal)?
2.5 gal
3.785 L
1 mL
–3
1 gal
2 sig. fig.s 4 sig. fig.s
1 cm3
10 L
1 mL
∞ sig. fig.s
∞ sig. fig.s
= 9462.5 cm3
9.5 x 103 cm3
Answer: _______________________
3. The Earth has a mass of 5.974 x 1021 metric tons (tonnes, t), what is this mass in pounds (lbs)?
5.974 x 1021 tonnes
4 sig. fig.s
1000 kg
2.205 lbs
1 tonne
1 kg
∞ sig. fig.s
4 sig. fig.s
1.317 x 1025 lbs
Answer: _______________________
= 1.317267 x 1025 lbs
4. A leaky faucet drips 1 drop of water every 1.2 seconds. How many gallons of water will this faucet
waste as a result during a month with exactly 31 days?
Note: 1 mL = 20. drops
31 days
24 hrs
60 min
60 sec
1 drop
1 mL
10-3 L
1 gal
1 day
1 hr
1 min
1.2 sec
20. drops
1 mL
3.785 L
2 sig. fig.s
2 sig. fig.s
∞ sig. fig.s
4 sig. fig.s
∞ sig. fig.s ∞ sig. fig.s ∞ sig. fig.s ∞ sig. fig.s
(exact)
= 29.48480845 gal
29 gal
Answer: _______________________
5. Platinum has a density of 21.090 g/cm3 and costs $2014 per troy ounce (oz t). At this price, what
volume (in cm3) of platinum could be purchased for exactly $150,000.00?
Note: 1 troy ounce (oz t) = 31.103 g
1 oz t
31.103 g
1 cm3
$2014
1 oz t
21.090 g
4 sig. fig.s
5 sig. fig.s
5 sig. fig.s
$150,000.00
∞ sig. fig.s
= 109.8392335 cm3
109.8 cm3
Answer: _______________________
Part II. The following problems require the use of equations and may also require dimensional
analysis. Show any equation used with variables, then again plugged in with numbers in the
same order as the variables.
6. What is the volume of the object to the right in cubic inches (in3)?
V=l×w×h
V = (3.4 ft) (1.2 ft) (1.8 ft) = 7.344 ft3
7.344 ft
3
kg
2 sig. fig.s
12 in
1 ft
height = 1.8 ft
width = 1.2 ft
length = 3.4 ft
volume = length × width × height
3
= 12690.432 in3
∞ sig. fig.s
1.3 x 104 in3
Answer: _______________________
Note: All available digits are used throughout
the calculation and only the final answer
is rounded to the correct number of
significant figures.
Cubing a conversion factor cubes all the
numbers and units inside.
7. What is a temperature of 350.0 Kelvin in degrees Fahrenheit?
K = °C + 273.15
°C = K – 273.15
°C = (350.0K) – 273.15 = 76.8 5°C
170. °F
Answer: _______________________
Note: You can consider the
1.8 and 32 in the °C to
°F equation to be exact
°F = 1.8 (°C) + 32
numbers (∞ sig. fig.s).
However, the 273.15 in
°F = 1.8 (76.85°C) + 32
the °C to K equation
°F = 138 .33 + 32
is only significant
to the 1/100 column.
°F = 170. 33°F