Dimensional Analysis Practice
Unit Conversion Exercises
This worksheet provides practice with unit conversions using dimensional analysis. Problems progress from basic single-step conversions to multi-step problems involving compound units. All solutions show proper significant figure tracking.
Basic Single-Step Conversions
Difficulty: Basic • Problems: 1–8 • Single conversion factor, all exact conversions
Convert 45.0 cm to meters.
Solution
The conversion 1 m = 100 cm is exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 45.\bar{0}~\text{cm} \left( \frac{1~\text{m}}{100~\text{cm}} \right) &= 0.45\bar{0}0~\text{m} \\[1.5ex] &= 0.450~\text{m} \end{align*} \]
Convert 2.8 km to meters.
Solution
The conversion 1 km = 103 m is exact. The answer is limited by the starting value (2 sig figs).
\[ \begin{align*} 2.\bar{8}~\text{km} \left( \frac{10^3~\text{m}}{1~\text{km}} \right) &= 2.\bar{8}00 \times 10^{3}~\text{m} \\[1.5ex] &= 2.8 \times 10^{3}~\text{m} \end{align*} \]
Convert 0.572 kg to grams.
Solution
The conversion 1 kg = 103 g is exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 0.57\bar{2}~\text{kg} \left( \frac{10^3~\text{g}}{1~\text{kg}} \right) &= 57\bar{2}.0~\text{g} \\[1.5ex] &= 572~\text{g} \end{align*} \]
Convert 825 mg to grams.
Solution
The conversion 1 g = 103 mg is exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 82\bar{5}~\text{mg} \left( \frac{1~\text{g}}{10^3~\text{mg}} \right) &= 0.82\bar{5}0~\text{g} \\[1.5ex] &= 0.825~\text{g} \end{align*} \]
Convert 3.50 L to milliliters.
Solution
The conversion 1 L = 103 mL is exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 3.5\bar{0}~\text{L} \left( \frac{10^3~\text{mL}}{1~\text{L}} \right) &= 3.5\bar{0}0 \times 10^{3}~\text{mL} \\[1.5ex] &= 3.50 \times 10^{3}~\text{mL} \end{align*} \]
Convert 7.5 hours to minutes.
Solution
The conversion 1 h = 60 min is exact. The answer is limited by the starting value (2 sig figs).
\[ \begin{align*} 7.\bar{5}~\text{h} \left( \frac{60~\text{min}}{1~\text{h}} \right) &= 4\bar{5}0~\text{min} \\[1.5ex] &= 4.5 \times 10^{2}~\text{min} \end{align*} \]
Convert 12.0 inches to centimeters.
Solution
The conversion 1 in = 2.54 cm is exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 12.\bar{0}~\text{in} \left( \frac{2.54~\text{cm}}{1~\text{in}} \right) &= 30.\bar{4}8~\text{cm} \\[1.5ex] &= 30.5~\text{cm} \end{align*} \]
Convert 5.00 gallons to liters. Use 1 gal = 3.785412 L.
Solution
The conversion factor 1 gal = 3.785412 L has 7 sig figs. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 5.0\bar{0}~\text{gal} \left( \frac{3.78541\bar{2}~\text{L}}{1~\text{gal}} \right) &= 18.\bar{9}27~\text{L} \\[1.5ex] &= 18.9~\text{L} \end{align*} \]
Multi-Step Conversions
Difficulty: Intermediate • Problems: 9–16 • Chained conversions, mixed unit systems
Convert 4.2 km to millimeters.
Solution
We chain two exact conversions: km → m → mm. The answer is limited by the starting value (2 sig figs).
\[ \begin{align*} 4.\bar{2}~\text{km} \left( \frac{10^3~\text{m}}{1~\text{km}} \right) \left( \frac{10^3~\text{mm}}{1~\text{m}} \right) &= 4.\bar{2}00 \times 10^{6}~\text{mm} \\[1.5ex] &= 4.2 \times 10^{6}~\text{mm} \end{align*} \]
Convert 0.0350 g to micrograms.
Solution
We chain two exact conversions: g → mg → μg, or use g → μg directly (1 g = 106 μg). The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 0.035\bar{0}~\text{g} \left( \frac{10^6~\upmu\text{g}}{1~\text{g}} \right) &= 3.5\bar{0}0 \times 10^{4}~\upmu\text{g} \\[1.5ex] &= 3.50 \times 10^{4}~\upmu\text{g} \end{align*} \]
Convert 8.5 μg to kilograms.
Solution
We use the exact conversion 1 kg = 109 μg. The answer is limited by the starting value (2 sig figs).
\[ \begin{align*} 8.\bar{5}~\upmu\text{g} \left( \frac{1~\text{kg}}{10^9~\upmu\text{g}} \right) &= 8.\bar{5}0 \times 10^{-9}~\text{kg} \\[1.5ex] &= 8.5 \times 10^{-9}~\text{kg} \end{align*} \]
Convert 2.50 miles to meters. Use 1 mi = 1.609344 km (exact).
Solution
Both conversions are exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 2.5\bar{0}~\text{mi} \left( \frac{1.609344~\text{km}}{1~\text{mi}} \right) \left( \frac{10^3~\text{m}}{1~\text{km}} \right) &= 4.0\bar{2}36 \times 10^{3}~\text{m} \\[1.5ex] &= 4.02 \times 10^{3}~\text{m} \end{align*} \]
Convert 16.0 oz to grams. Use 1 lb = 16 oz (exact) and 1 lb = 453.59237 g (exact).
Solution
Both conversions are exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 16.\bar{0}~\text{oz} \left( \frac{1~\text{lb}}{16~\text{oz}} \right) \left( \frac{453.59237~\text{g}}{1~\text{lb}} \right) &= 45\bar{3}.59~\text{g} \\[1.5ex] &= 454~\text{g} \end{align*} \]
Convert 250. mL to gallons. Use 1 gal = 3.785412 L.
Solution
The metric conversion is exact. The conversion factor 1 gal = 3.785412 L has 7 sig figs. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 25\bar{0}.~\text{mL} \left( \frac{1~\text{L}}{10^3~\text{mL}} \right) \left( \frac{1~\text{gal}}{3.78541\bar{2}~\text{L}} \right) &= 0.066\bar{0}43~\text{gal} \\[1.5ex] &= 0.0660~\text{gal} \end{align*} \]
Convert 3.00 days to seconds.
Solution
All time conversions are exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} 3.0\bar{0}~\text{d} \left( \frac{24~\text{h}}{1~\text{d}} \right) \left( \frac{3600~\text{s}}{1~\text{h}} \right) &= 2.5\bar{9}2 \times 10^{5}~\text{s} \\[1.5ex] &= 2.59 \times 10^{5}~\text{s} \end{align*} \]
Convert 45 nm to cm.
Solution
All metric conversions are exact. The answer is limited by the starting value (2 sig figs).
\[ \begin{align*} 4\bar{5}~\text{nm} \left( \frac{10^{-9}~\text{m}}{1~\text{nm}} \right) \left( \frac{100~\text{cm}}{1~\text{m}} \right) &= 4.\bar{5}0 \times 10^{-6}~\text{cm} \\[1.5ex] &= 4.5 \times 10^{-6}~\text{cm} \end{align*} \]
Compound Unit Conversions
Difficulty: Intermediate • Problems: 17–22 • Speed, density, and flow rate conversions
Convert 55 mph to m s−1. Use 1 mi = 1.609344 km (exact).
Solution
All conversions are exact. The answer is limited by the starting value (2 sig figs).
\[ \begin{align*} \frac{5\bar{5}~\text{mi}}{\text{h}} \left( \frac{1.609344~\text{km}}{1~\text{mi}} \right) \left( \frac{10^3~\text{m}}{1~\text{km}} \right) \left( \frac{1~\text{h}}{3600~\text{s}} \right) &= 2\bar{4}.58~\text{m}~\text{s}^{-1} \\[1.5ex] &= 25~\text{m}~\text{s}^{-1} \end{align*} \]
Convert 100. km h−1 to m s−1.
Solution
All conversions are exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} \frac{10\bar{0}.~\text{km}}{\text{h}} \left( \frac{10^3~\text{m}}{1~\text{km}} \right) \left( \frac{1~\text{h}}{3600~\text{s}} \right) &= 27.\bar{7}77~\text{m}~\text{s}^{-1} \\[1.5ex] &= 27.8~\text{m}~\text{s}^{-1} \end{align*} \]
Convert 1.05 g mL−1 to kg L−1.
Solution
All conversions are exact. Note that (1 g/mL) × (1 kg/103 g) × (103 mL/L) = 1 kg/L, so the numerical value is unchanged. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} \frac{1.0\bar{5}~\text{g}}{\text{mL}} \left( \frac{1~\text{kg}}{10^3~\text{g}} \right) \left( \frac{10^3~\text{mL}}{1~\text{L}} \right) &= 1.0\bar{5}0~\text{kg}~\text{L}^{-1} \\[1.5ex] &= 1.05~\text{kg}~\text{L}^{-1} \end{align*} \]
Convert 2.70 g cm−3 to kg m−3.
Solution
All conversions are exact. Note that cm3 → m3 requires cubing the conversion factor: (100 cm/m)3 = 106 cm3/m3. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} \frac{2.7\bar{0}~\text{g}}{\text{cm}^3} \left( \frac{1~\text{kg}}{10^3~\text{g}} \right) \left( \frac{10^6~\text{cm}^3}{1~\text{m}^3} \right) &= 2.7\bar{0}0 \times 10^{3}~\text{kg}~\text{m}^{-3} \\[1.5ex] &= 2.70 \times 10^{3}~\text{kg}~\text{m}^{-3} \end{align*} \]
Convert 4.50 L min−1 to mL s−1.
Solution
All conversions are exact. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} \frac{4.5\bar{0}~\text{L}}{\text{min}} \left( \frac{10^3~\text{mL}}{1~\text{L}} \right) \left( \frac{1~\text{min}}{60~\text{s}} \right) &= 7\bar{5}.00~\text{mL}~\text{s}^{-1} \\[1.5ex] &= 75.0~\text{mL}~\text{s}^{-1} \end{align*} \]
Convert 62.4 lb ft−3 to g cm−3. Use 1 lb = 453.59237 g (exact) and 1 ft = 12 in (exact), 1 in = 2.54 cm (exact).
Solution
All conversions are exact. Note: 1 ft = 12 × 2.54 cm = 30.48 cm, so (1 ft)3 = (30.48 cm)3 = 28316.846592 cm3. The answer is limited by the starting value (3 sig figs).
\[ \begin{align*} \frac{62.\bar{4}~\text{lb}}{\text{ft}^3} \left( \frac{453.59237~\text{g}}{1~\text{lb}} \right) \left( \frac{1~\text{ft}^3}{28316.846592~\text{cm}^3} \right) &= 0.99\bar{9}76~\text{g}~\text{cm}^{-3} \\[1.5ex] &= 1.00~\text{g}~\text{cm}^{-3} \end{align*} \]
Applied and Challenge Problems
Difficulty: Challenge • Problems: 23–28 • Multi-step real-world applications
A runner completes a 10.0 km race in 42 min 15 s. What was their average speed in m s−1?
Solution
First, convert time to seconds: 42 min 15 s = (42 × 60) + 15 = 2535 s (exact, counted).
The distance 10.0 km has 3 sig figs and limits the answer.
\[ \begin{align*} \frac{10.\bar{0}~\text{km}}{2535~\text{s}} \left( \frac{10^3~\text{m}}{1~\text{km}} \right) &= 3.9\bar{4}48~\text{m}~\text{s}^{-1} \\[1.5ex] &= 3.94~\text{m}~\text{s}^{-1} \end{align*} \]
A medication dosage is 15 mg per kg of body weight. How many grams of medication should be administered to a patient weighing 165 lb? Use 1 lb = 453.59237 g (exact).
Solution
The dosage (15 mg/kg) has 2 sig figs. The body weight (165 lb) has 3 sig figs. The answer is limited to 2 sig figs.
\[ \begin{align*} 16\bar{5}~\text{lb} \left( \frac{453.59237~\text{g}}{1~\text{lb}} \right) \left( \frac{1~\text{kg}}{10^3~\text{g}} \right) \left( \frac{1\bar{5}~\text{mg}}{1~\text{kg}} \right) \left( \frac{1~\text{g}}{10^3~\text{mg}} \right) &= 1.\bar{1}23~\text{g} \\[1.5ex] &= 1.1~\text{g} \end{align*} \]
Gasoline costs $3.459 per gallon. What is the cost in dollars per liter? Use 1 gal = 3.785412 L.
Solution
The price ($3.459/gal) has 4 sig figs. The conversion factor has 7 sig figs. The answer is limited by the price (4 sig figs).
\[ \begin{align*} \frac{\$3.45\bar{9}}{1~\text{gal}} \left( \frac{1~\text{gal}}{3.78541\bar{2}~\text{L}} \right) &= \$0.913\bar{6}4~\text{L}^{-1} \\[1.5ex] &= \$0.9136~\text{L}^{-1} \end{align*} \]
A water pipe delivers 2.5 gal min−1. How many liters does it deliver in 1.0 hour? Use 1 gal = 3.785412 L.
Solution
The flow rate (2.5 gal/min) has 2 sig figs. The conversion factor has 7 sig figs. The time conversion is exact. The answer is limited by the flow rate (2 sig figs).
\[ \begin{align*} \frac{2.\bar{5}~\text{gal}}{\text{min}} \left( \frac{3.78541\bar{2}~\text{L}}{1~\text{gal}} \right) \left( \frac{60~\text{min}}{1~\text{h}} \right) \left( 1.0~\text{h} \right) &= 5.\bar{6}78 \times 10^{2}~\text{L} \\[1.5ex] &= 5.7 \times 10^{2}~\text{L} \end{align*} \]
The density of gold is 19.3 g cm−3. What is the mass in kg of a gold bar with dimensions 25.0 cm × 8.00 cm × 4.00 cm?
Solution
First, calculate the volume. The dimensions have 3 sig figs each.
\[ \begin{align*} V &= 25.\bar{0}~\text{cm} \times 8.0\bar{0}~\text{cm} \times 4.0\bar{0}~\text{cm} \\[1.5ex] &= 80\bar{0}.0~\text{cm}^3 \end{align*} \]
The density (19.3 g/cm3) has 3 sig figs. The answer is limited to 3 sig figs.
\[ \begin{align*} 80\bar{0}.0~\text{cm}^3 \left( \frac{19.\bar{3}~\text{g}}{1~\text{cm}^3} \right) \left( \frac{1~\text{kg}}{10^3~\text{g}} \right) &= 15.\bar{4}4~\text{kg} \\[1.5ex] &= 15.4~\text{kg} \end{align*} \]
Light travels at 2.998 × 108 m s−1. How many miles does light travel in one year (365.25 days)? Express your answer in scientific notation.
Solution
The speed of light has 4 sig figs and limits the answer. The time conversions and 1 mi = 1.609344 km are exact. The value 365.25 days is the Julian year, which is exact by definition.
\[ \begin{align*} \frac{2.99\bar{8} \times 10^{8}~\text{m}}{\text{s}} \left( \frac{1~\text{km}}{10^3~\text{m}} \right) \left( \frac{1~\text{mi}}{1.609344~\text{km}} \right) \left( \frac{3600~\text{s}}{1~\text{h}} \right) \left( \frac{24~\text{h}}{1~\text{d}} \right) \left( 365.25~\text{d} \right) &= 5.87\bar{8}5 \times 10^{12}~\text{mi} \\[1.5ex] &= 5.878 \times 10^{12}~\text{mi} \end{align*} \]
This distance is called one light-year.