10/29 Practice Problems - Answer Key

Problem 1

In your own words, explain what density is and what it tells us about a substance. Why do different substances have different densities?

Answer:

Density is the amount of mass packed into a given volume of a substance. It's calculated using the formula: Density = mass/volume (D = m/V). Density tells us how tightly packed the particles of a substance are.

Different substances have different densities because:

• Atomic/molecular mass: Heavier atoms or molecules contribute more mass to the same volume
• Particle arrangement: How tightly the atoms or molecules are packed together affects density
• State of matter: Solids are generally denser than liquids, which are denser than gases, because particles are more tightly packed in solids

Problem 2

You have two blocks of the same size and shape. One is made of aluminum and one is made of lead. The lead block feels much heavier. Explain this observation using the concept of density.

Answer:

Since both blocks have the same volume (same size and shape), the difference in weight is due to their different densities.

Lead has a much higher density than aluminum:

• Density of aluminum ≈ 2.70 g/cm³
• Density of lead ≈ 11.3 g/cm³

Using the formula D = m/V, we can rearrange to get m = D × V.

Since volume (V) is the same for both blocks, but lead's density is about 4 times greater than aluminum's density, the lead block will have about 4 times more mass. More mass means more weight, which is why the lead block feels much heavier.

Problem 3

Describe the water displacement method for measuring the volume of an irregular solid object (like a rock). Include the steps you would follow and explain why this method works.

Answer:

Steps for water displacement method:

1. Fill a graduated cylinder partially with water (enough to cover the object, but with room for the water level to rise)
2. Record the initial water level (V₁)
3. Carefully place the irregular solid object into the water, making sure it's completely submerged
4. Record the new water level (V₂)
5. Calculate the volume of the object: Volume = V₂ - V₁

Why this method works:

When you submerge an object in water, it displaces (pushes aside) a volume of water equal to its own volume. The water level rises by an amount exactly equal to the volume of the object. By measuring how much the water level rises, we can determine the volume of the irregular object, even though we couldn't measure it directly with a ruler.

Problem 4

What is specific heat? How does water's high specific heat capacity affect Earth's climate and weather patterns? Give at least one example.

Answer:

What is specific heat?

Specific heat (c) is the amount of heat energy required to raise the temperature of 1 gram of a substance by 1°C (or 1 K). It's measured in J/g°C. Different substances have different specific heats - some materials heat up or cool down quickly (low specific heat), while others resist temperature changes (high specific heat).

Water's high specific heat (4.18 J/g°C):

Water has one of the highest specific heats of common substances, which means it takes a lot of energy to change its temperature.

Effects on Earth's climate and weather:

• Coastal climates are moderate: Oceans absorb huge amounts of heat during the day/summer without getting very hot, then release that heat slowly at night/winter. This keeps coastal areas cooler in summer and warmer in winter compared to inland areas.
• Ocean currents regulate global climate: Warm ocean currents carry heat from the equator toward the poles, while cold currents move cool water toward the equator, helping to distribute heat around the planet.
• Weather patterns: Large bodies of water influence local weather by moderating temperature changes and affecting humidity and precipitation patterns.
• Daily temperature swings: Areas near water have smaller differences between day and night temperatures because water heats up and cools down slowly.

Problem 5

Two identical pots of water are heated on a stove. Pot A contains 1 L of water and Pot B contains 2 L of water. Both start at the same temperature and receive the same amount of heat energy. Which pot will have a higher final temperature? Explain your reasoning using the concept of specific heat.

Answer:

Pot A (1 L) will have the higher final temperature.

Reasoning using the heat equation:

The heat equation is: q = mcΔT

Where:

• q = heat energy absorbed
• m = mass
• c = specific heat
• ΔT = change in temperature

We can rearrange this to find ΔT:

\[\Delta T = \frac{q}{mc}\]

Analysis:

• Both pots receive the same q (heat energy)
• Both contain water, so c is the same (4.18 J/g°C)
• Pot A has less mass (m) than Pot B (1 L vs 2 L of water)

Since ΔT = q/(mc), and m is in the denominator, a smaller mass results in a larger ΔT.

Pot A has half the mass of Pot B, so with the same heat energy input, Pot A will experience twice the temperature increase!

Problem 6

Express 0.000456 in scientific notation with the correct number of significant figures.

Solution:

Step 1: Move the decimal point to create a number between 1 and 10

Move the decimal 4 places to the right: 4.56

Step 2: Count the significant figures

0.000456 has 3 significant figures (4, 5, and 6)

Step 3: Write in scientific notation

Since we moved the decimal 4 places to the right, the exponent is -4.

Problem 7

How many significant figures are in the measurement 3050.0 g?

Solution:

Step 1: Apply significant figure rules

- All non-zero digits are significant: 3, 5
- Zeros between non-zero digits are significant: 0 (between 3 and 5)
- Trailing zeros after a decimal point are significant: 0.0

Step 2: Count them

3, 0, 5, 0, 0 = 5 significant figures

Problem 8

Convert 2.5 L to mL.

Solution:

Step 1: Use the conversion factor

1 L = 1000 mL

Step 2: Set up the conversion

\[2.5 \cancel{\text{ L}} \times \frac{1000 \text{ mL}}{1 \cancel{\text{ L}}} = 2500 \text{ mL}\]

Step 3: Apply significant figures

2.5 has 2 sig figs, so answer should have 2 sig figs: 2500 = \(2.5 \times 10^3\) mL

Problem 9

Convert 350 cm³ to mL.

Solution:

Step 1: Use the conversion factor

1 cm³ = 1 mL (this is a direct equivalence)

Step 2: Convert

\[350 \text{ cm}^3 \times \frac{1 \text{ mL}}{1 \text{ cm}^3} = 350 \text{ mL}\]

Problem 10

Convert 75°F to °C. (Show your work and formula.)

Solution:

Step 1: Write the formula

\[°C = \frac{5}{9}(°F - 32)\]

Step 2: Substitute and solve

\[°C = \frac{5}{9}(75 - 32)\]

\[°C = \frac{5}{9}(43)\]

\[°C = 23.89°C\]

Step 3: Apply significant figures

75 has 2 sig figs, so round to 2 sig figs: 24°C

Problem 11

Convert 298 K to °C.

Solution:

Step 1: Write the formula

\[°C = K - 273.15\]

Step 2: Substitute and solve

\[°C = 298 - 273.15 = 24.85°C\]

Step 3: Apply significant figures

298 has 3 sig figs, so answer is 24.9°C (3 sig figs)

Problem 12

Convert 3.5 km to m.

Solution:

Step 1: Use the conversion factor

1 km = 1000 m

Step 2: Convert

\[3.5 \cancel{\text{ km}} \times \frac{1000 \text{ m}}{1 \cancel{\text{ km}}} = 3500 \text{ m}\]

Step 3: Apply significant figures

3.5 has 2 sig figs: \(3.5 \times 10^3\) m

Problem 13

A sample of aluminum has a mass of 54.0 g and a volume of 20.0 cm³. Calculate the density of aluminum.

Solution:

Step 1: Write the density formula

\[\text{Density} = \frac{\text{mass}}{\text{volume}}\]

Step 2: Substitute values

\[\text{Density} = \frac{54.0 \text{ g}}{20.0 \text{ cm}^3} = 2.70 \text{ g/cm}^3\]

Problem 14

Calculate \(\Delta T\) (change in temperature) when water is heated from 25°C to 88°C.

Solution:

Step 1: Write the formula

\[\Delta T = T_{\text{final}} - T_{\text{initial}}\]

Step 2: Substitute and solve

\[\Delta T = 88°C - 25°C = 63°C\]

Problem 15

How much heat energy is required to raise the temperature of 100 g of water by 15°C? (\(c_{\text{water}} = 4.18\) J/g°C)

Solution:

Step 1: Write the heat equation

\[q = mc\Delta T\]

Step 2: Identify values

m = 100 g
c = 4.18 J/g°C
ΔT = 15°C

Step 3: Calculate

\[q = (100 \text{ g})(4.18 \text{ J/g°C})(15°C) = 6270 \text{ J}\]

Step 4: Apply significant figures

15 has 2 sig figs (limiting): 6300 J or 6.3 kJ

Problem 16

A chemist measures the mass of a chemical sample as \(3.45 \times 10^{-3}\) kg. Express this mass in grams using scientific notation.

Solution:

Step 1: Convert kg to g

1 kg = 1000 g

\[3.45 \times 10^{-3} \cancel{\text{ kg}} \times \frac{1000 \text{ g}}{1 \cancel{\text{ kg}}} = 3.45 \text{ g}\]

Step 2: Express in proper scientific notation

\(3.45 \times 10^{-3} \times 10^3 = 3.45 \times 10^0 = 3.45\) g

Or keep in scientific notation: \(3.45 \times 10^0\) g

Problem 17

A student records the length of a pencil as 18.50 cm. She then multiplies this by 2.1 to calculate something. What is the answer with the correct number of significant figures?

Solution:

Step 1: Multiply the numbers

\[18.50 \times 2.1 = 38.85\]

Step 2: Apply significant figure rules for multiplication

18.50 has 4 sig figs
2.1 has 2 sig figs
Answer should have 2 sig figs (the smaller number)

Step 3: Round to 2 sig figs

38.85 → 39

Problem 18

A swimming pool contains 85,000 L of water. Convert this volume to mL and express your answer in scientific notation.

Solution:

Step 1: Convert L to mL

1 L = 1000 mL

\[85,000 \cancel{\text{ L}} \times \frac{1000 \text{ mL}}{1 \cancel{\text{ L}}} = 85,000,000 \text{ mL}\]

Step 2: Express in scientific notation

\[85,000,000 \text{ mL} = 8.5 \times 10^7 \text{ mL}\]

Problem 19

A rectangular aquarium measures 50.0 cm long, 30.0 cm wide, and 40.0 cm tall. Calculate the volume in cm³, then convert to liters.

Solution:

Step 1: Calculate volume in cm³

\[\text{Volume} = \text{length} \times \text{width} \times \text{height}\]

\[\text{Volume} = 50.0 \text{ cm} \times 30.0 \text{ cm} \times 40.0 \text{ cm} = 60,000 \text{ cm}^3\]

Step 2: Convert cm³ to L

1 L = 1000 cm³

\[60,000 \cancel{\text{ cm}^3} \times \frac{1 \text{ L}}{1000 \cancel{\text{ cm}^3}} = 60.0 \text{ L}\]

Problem 20

A weather forecast predicts a high temperature of 86°F. The student wants to know if this is warmer than 30°C. Convert 86°F to Celsius and determine which temperature is warmer.

Solution:

Step 1: Write the formula

\[°C = \frac{5}{9}(°F - 32)\]

Step 2: Substitute and solve

\[°C = \frac{5}{9}(86 - 32)\]

\[°C = \frac{5}{9}(54)\]

\[°C = 30°C\]

Step 3: Compare

86°F = 30°C, so they are the same temperature!

Problem 21

Liquid nitrogen boils at 77 K. Convert this temperature to both Celsius and Fahrenheit.

Solution:

Step 1: Convert K to °C

\[°C = K - 273.15\]

\[°C = 77 - 273.15 = -196.15°C\]

With sig figs (2): -196°C or -2.0 × 10² °C

Step 2: Convert °C to °F

\[°F = \frac{9}{5}(°C) + 32\]

\[°F = \frac{9}{5}(-196) + 32\]

\[°F = -352.8 + 32 = -320.8°F\]

With sig figs (2): -3.2 × 10² °F

Problem 22

A marathon runner runs 42.195 km. Convert this distance to meters, then convert to millimeters.

Solution:

Step 1: Convert km to m

1 km = 1000 m

\[42.195 \cancel{\text{ km}} \times \frac{1000 \text{ m}}{1 \cancel{\text{ km}}} = 42,195 \text{ m}\]

Step 2: Convert m to mm

1 m = 1000 mm

\[42,195 \cancel{\text{ m}} \times \frac{1000 \text{ mm}}{1 \cancel{\text{ m}}} = 42,195,000 \text{ mm}\]

Step 3: Express in scientific notation

\[4.2195 \times 10^7 \text{ mm}\]

Problem 23

A gold bar has a mass of 386 g and a volume of 20.0 cm³. Calculate the density of gold in g/cm³. Then, determine the volume of a 50.0 g gold nugget with the same density.

Solution:

Step 1: Calculate density

\[\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{386 \text{ g}}{20.0 \text{ cm}^3} = 19.3 \text{ g/cm}^3\]

Step 2: Find volume of 50.0 g nugget

Rearrange density formula: \(\text{Volume} = \frac{\text{mass}}{\text{density}}\)

\[\text{Volume} = \frac{50.0 \text{ g}}{19.3 \text{ g/cm}^3} = 2.59 \text{ cm}^3\]

Problem 24

An unknown liquid has a density of 0.85 g/mL. If you have 250 mL of this liquid, what is its mass in grams? If this liquid is then poured into a container that can hold 500 cm³, will it overflow?

Solution:

Step 1: Calculate mass

Rearrange density: \(\text{mass} = \text{density} \times \text{volume}\)

\[\text{mass} = 0.85 \text{ g/mL} \times 250 \text{ mL} = 212.5 \text{ g}\]

With sig figs (2): 210 g or 2.1 × 10² g

Step 2: Determine if it will overflow

Remember: 1 mL = 1 cm³, so 500 cm³ = 500 mL

We have 250 mL of liquid and the container holds 500 mL.

250 mL < 500 mL, so it will NOT overflow.

Problem 25

A 150 g sample of iron (\(c_{\text{iron}} = 0.449\) J/g°C) is heated from 25°C to 125°C. Calculate the amount of heat energy absorbed. If this same amount of heat energy were used to heat 150 g of water (\(c_{\text{water}} = 4.18\) J/g°C) starting at 25°C, what would be the final temperature of the water?

Solution:

Part 1: Heat absorbed by iron

Step 1: Identify values

m = 150 g
c = 0.449 J/g°C
ΔT = 125°C - 25°C = 100°C

Step 2: Calculate heat

\[q = mc\Delta T = (150 \text{ g})(0.449 \text{ J/g°C})(100°C)\]

\[q = 6735 \text{ J} = 6.74 \times 10^3 \text{ J}\]

With sig figs (3 from 150): \(6.74 \times 10^3\) J or 6.74 kJ

Part 2: Final temperature of water

Step 3: Use same heat for water

\[q = mc\Delta T\]

\[6735 = (150)(4.18)\Delta T\]

\[6735 = 627\Delta T\]

\[\Delta T = \frac{6735}{627} = 10.7°C\]

Step 4: Calculate final temperature

\[T_{\text{final}} = T_{\text{initial}} + \Delta T = 25°C + 10.7°C = 35.7°C\]

With sig figs: 36°C

Key insight: Water has a much higher specific heat than iron (4.18 vs 0.449), so it takes the same energy to raise water's temperature by only ~11°C versus raising iron's temperature by 100°C!

Problem 26

Calculate the molar mass of water (H₂O). (H = 1.01 g/mol, O = 16.00 g/mol)

Solution:

Step 1: Identify the number of each atom in the formula

H₂O contains:

• 2 hydrogen atoms (H)
• 1 oxygen atom (O)

Step 2: Multiply each atom's molar mass by the number of atoms

H: 2 × 1.01 g/mol = 2.02 g/mol
O: 1 × 16.00 g/mol = 16.00 g/mol

Step 3: Add them together

Molar mass of H₂O = 2.02 + 16.00 = 18.02 g/mol

Problem 27

Calculate the molar mass of carbon dioxide (CO₂). (C = 12.01 g/mol, O = 16.00 g/mol)

Solution:

Step 1: Identify the number of each atom in the formula

CO₂ contains:

• 1 carbon atom (C)
• 2 oxygen atoms (O)

Step 2: Multiply each atom's molar mass by the number of atoms

C: 1 × 12.01 g/mol = 12.01 g/mol
O: 2 × 16.00 g/mol = 32.00 g/mol

Step 3: Add them together

Molar mass of CO₂ = 12.01 + 32.00 = 44.01 g/mol

Problem 28

Calculate the molar mass of sodium chloride (NaCl). (Na = 22.99 g/mol, Cl = 35.45 g/mol)

Solution:

Step 1: Identify the number of each atom in the formula

NaCl contains:

• 1 sodium atom (Na)
• 1 chlorine atom (Cl)

Step 2: Multiply each atom's molar mass by the number of atoms

Na: 1 × 22.99 g/mol = 22.99 g/mol
Cl: 1 × 35.45 g/mol = 35.45 g/mol

Step 3: Add them together

Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol

Problem 29

Calculate the molar mass of glucose (C₆H₁₂O₆). (C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol)

Solution:

Step 1: Identify the number of each atom in the formula

C₆H₁₂O₆ contains:

• 6 carbon atoms (C)
• 12 hydrogen atoms (H)
• 6 oxygen atoms (O)

Step 2: Multiply each atom's molar mass by the number of atoms

C: 6 × 12.01 g/mol = 72.06 g/mol
H: 12 × 1.01 g/mol = 12.12 g/mol
O: 6 × 16.00 g/mol = 96.00 g/mol

Step 3: Add them together

Molar mass of C₆H₁₂O₆ = 72.06 + 12.12 + 96.00 = 180.18 g/mol

Problem 30

Calculate the molar mass of calcium carbonate (CaCO₃). (Ca = 40.08 g/mol, C = 12.01 g/mol, O = 16.00 g/mol)

Solution:

Step 1: Identify the number of each atom in the formula

CaCO₃ contains:

• 1 calcium atom (Ca)
• 1 carbon atom (C)
• 3 oxygen atoms (O)

Step 2: Multiply each atom's molar mass by the number of atoms

Ca: 1 × 40.08 g/mol = 40.08 g/mol
C: 1 × 12.01 g/mol = 12.01 g/mol
O: 3 × 16.00 g/mol = 48.00 g/mol

Step 3: Add them together

Molar mass of CaCO₃ = 40.08 + 12.01 + 48.00 = 100.09 g/mol

Problem 31

How many moles are in 88.0 g of carbon dioxide (CO₂)? (Molar mass of CO₂ = 44.01 g/mol)

Solution:

Step 1: Write the conversion from grams to moles

Use the molar mass as a conversion factor:

\[\text{moles} = \frac{\text{grams}}{\text{molar mass}}\]

Step 2: Set up the calculation with units

\[88.0 \cancel{\text{ g CO}_2} \times \frac{1 \text{ mol CO}_2}{44.01 \cancel{\text{ g CO}_2}}\]

Step 3: Calculate

\[\frac{88.0}{44.01} = 2.00 \text{ mol CO}_2\]

Problem 32

How many molecules are in 36.0 g of water (H₂O)? (Molar mass of H₂O = 18.02 g/mol, Avogadro's number = \(6.022 \times 10^{23}\))

Solution:

Step 1: Convert grams to moles

\[36.0 \cancel{\text{ g H}_2\text{O}} \times \frac{1 \text{ mol H}_2\text{O}}{18.02 \cancel{\text{ g H}_2\text{O}}} = 1.998 \text{ mol H}_2\text{O}\]

Step 2: Convert moles to molecules using Avogadro's number

\[1.998 \cancel{\text{ mol H}_2\text{O}} \times \frac{6.022 \times 10^{23} \text{ molecules}}{1 \cancel{\text{ mol}}} = 1.203 \times 10^{24} \text{ molecules}\]

Step 3: Apply significant figures

36.0 has 3 sig figs, so answer should have 3 sig figs

Problem 33

A sample contains 117 g of sodium chloride (NaCl). Calculate (a) the number of moles of NaCl and (b) the number of formula units of NaCl in the sample. (Molar mass of NaCl = 58.44 g/mol, Avogadro's number = \(6.022 \times 10^{23}\))

Solution:

Part (a): Convert grams to moles

Step 1: Set up the conversion

\[117 \cancel{\text{ g NaCl}} \times \frac{1 \text{ mol NaCl}}{58.44 \cancel{\text{ g NaCl}}}\]

Step 2: Calculate

\[\frac{117}{58.44} = 2.002 \text{ mol NaCl}\]

Step 3: Apply significant figures

117 has 3 sig figs: 2.00 mol

Part (b): Convert moles to formula units

Step 4: Use Avogadro's number

\[2.00 \cancel{\text{ mol NaCl}} \times \frac{6.022 \times 10^{23} \text{ formula units}}{1 \cancel{\text{ mol}}} = 1.204 \times 10^{24} \text{ formula units}\]

Step 5: Apply significant figures

3 sig figs: \(1.20 \times 10^{24}\)

Note: For ionic compounds like NaCl, we use "formula units" instead of "molecules" because ionic compounds don't form discrete molecules.

Problem 34

A jewelry manufacturer produces sterling silver earrings. Sterling silver is 92.5% silver by mass. If the manufacturer needs to produce 250 pairs of earrings, and each earring has a mass of 3.2 grams, how many kilograms of pure silver are needed? Express your answer in scientific notation.

Solution:

Step 1: Calculate total number of earrings

250 pairs × 2 earrings/pair = 500 earrings

Step 2: Calculate total mass of sterling silver needed

500 earrings × 3.2 g/earring = 1600 g sterling silver

Step 3: Calculate mass of pure silver using percentage

92.5% = 0.925 (as a decimal)

Mass of pure silver = 0.925 × 1600 g = 1480 g

Step 4: Convert grams to kilograms

\[1480 \cancel{\text{ g}} \times \frac{1 \text{ kg}}{1000 \cancel{\text{ g}}} = 1.48 \text{ kg}\]

Step 5: Express in scientific notation with proper sig figs

Original data has 2 sig figs (3.2 g), so: \(1.5 \times 10^0\) kg or 1.5 kg

Key concepts used: Counting pairs, mass multiplication, percentage calculation, unit conversion (g → kg), scientific notation

Problem 35

A pharmaceutical company produces a cough syrup that is 12% active ingredient by volume. The syrup is sold in 4.0 fluid ounce bottles. If the company manufactures 5000 bottles per day, how many liters of active ingredient are used per day? (1 fluid ounce = 29.57 mL)

Solution:

Step 1: Calculate total volume of syrup per day in fluid ounces

5000 bottles × 4.0 fl oz/bottle = 20,000 fl oz total syrup

Step 2: Convert fluid ounces to milliliters

\[20,000 \cancel{\text{ fl oz}} \times \frac{29.57 \text{ mL}}{1 \cancel{\text{ fl oz}}} = 591,400 \text{ mL}\]

Step 3: Calculate volume of active ingredient using percentage

12% = 0.12 (as a decimal)

Volume of active ingredient = 0.12 × 591,400 mL = 70,968 mL

Step 4: Convert milliliters to liters

\[70,968 \cancel{\text{ mL}} \times \frac{1 \text{ L}}{1000 \cancel{\text{ mL}}} = 70.968 \text{ L}\]

Step 5: Apply significant figures

Limited by 4.0 (2 sig figs) and 12% (2 sig figs): 71 L or \(7.1 \times 10^1\) L

Key concepts used: Bottle counting, volume conversion (fl oz → mL → L), percentage by volume calculation

Problem 36

A chemist needs to determine the mass of a sample of ethanol (C₂H₅OH). She fills a graduated cylinder to the 50.0 mL mark, then adds an irregular piece of metal. The new volume reading is 65.3 mL. The metal has a density of 7.87 g/cm³. She then removes the metal and fills the cylinder with ethanol to the 50.0 mL mark. What is the mass of the ethanol in grams? (Density of ethanol = 0.789 g/mL, remember: 1 cm³ = 1 mL)

Solution:

Understanding the problem: This problem describes the water displacement method being used to measure metal volume. The key is finding the volume of ethanol, which is 50.0 mL.

Step 1: Determine volume of metal (using water displacement)

Volume of metal = Final reading - Initial reading

Volume of metal = 65.3 mL - 50.0 mL = 15.3 mL = 15.3 cm³

Note: This information about the metal is extra information - it's not needed to solve for ethanol mass!

Step 2: Identify volume of ethanol

The cylinder is filled to the 50.0 mL mark with ethanol.

Volume of ethanol = 50.0 mL

Step 3: Use density to calculate mass of ethanol

Density = mass/volume, so mass = density × volume

\[\text{mass} = 0.789 \frac{\text{g}}{\cancel{\text{mL}}} \times 50.0 \cancel{\text{ mL}} = 39.45 \text{ g}\]

Step 4: Apply significant figures

50.0 mL has 3 sig figs, 0.789 g/mL has 3 sig figs

Answer: 39.5 g (3 sig figs)

Key concepts used: Water displacement method (as context), density formula (D = m/V rearranged to m = D × V), unit equivalence (cm³ = mL)

Teaching note: This problem includes extra information about the metal to test whether students can identify the relevant data. The metal's volume and density are not needed to find the ethanol's mass!

Problem 37

An Olympic-sized swimming pool measures 50.0 meters long, 25.0 meters wide, and 2.00 meters deep. If the pool is filled to 95% capacity with water, what is the mass of the water in pounds? (Density of water = 1.00 g/mL, 1 m = 100 cm, 1 pound = 453.6 g)

Solution:

Step 1: Calculate volume of pool in cubic meters

Volume = length × width × depth

Volume = 50.0 m × 25.0 m × 2.00 m = 2500 m³

Step 2: Apply percentage to find actual volume of water

95% = 0.95 (as a decimal)

Volume of water = 0.95 × 2500 m³ = 2375 m³

Step 3: Convert cubic meters to cubic centimeters

1 m = 100 cm, so 1 m³ = (100 cm)³ = 1,000,000 cm³ = 10⁶ cm³

\[2375 \cancel{\text{ m}^3} \times \frac{10^6 \text{ cm}^3}{1 \cancel{\text{ m}^3}} = 2.375 \times 10^9 \text{ cm}^3\]

Step 4: Convert cm³ to mL (they're equal!)

1 cm³ = 1 mL

Volume = \(2.375 \times 10^9\) mL

Step 5: Use density to calculate mass in grams

mass = density × volume

\[\text{mass} = 1.00 \frac{\text{g}}{\cancel{\text{mL}}} \times 2.375 \times 10^9 \cancel{\text{ mL}} = 2.375 \times 10^9 \text{ g}\]

Step 6: Convert grams to pounds

\[2.375 \times 10^9 \cancel{\text{ g}} \times \frac{1 \text{ lb}}{453.6 \cancel{\text{ g}}} = 5.236 \times 10^6 \text{ lb}\]

Step 7: Apply significant figures

2.00 m has 3 sig figs (limiting): \(5.24 \times 10^6\) lb

Key concepts used: Volume calculation (L × W × H), percentage calculation, multi-step unit conversion (m³ → cm³ → mL), density to find mass (m = D × V), final conversion (g → lb)

Real-world context: An Olympic pool contains over 5 million pounds of water - that's equivalent to about 2,600 cars! This demonstrates why pool foundations must be extremely strong.