What is its density? A block of aluminum occupies a volume of 15.0 mL and weighs 40.5 g. 2) Mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. The mercury used to fill the cylinder weighs 306.0 g. Calculate the density of mercury. 3) What is the weight of the ethyl alcohol that exactly fills a 200.0 mL container? The density of ethyl alcohol is 0.789 g/mL. 4) A rectangular block of copper metal weighs 1896 g and has dimensions of 8.4 cm by 5.5 cm by 4.6 cm. Calculate the density of copper. 5) A flask that weighs 345.8 g is filled with 225 mL of carbon tetrachloride. The weight of the flask and carbon tetrachloride is found to be 703.55 g. Calculate the density of carbon tetrachloride. 6) Calculate the density of sulfuric acid if 35.4 mL of the acid weighs 65.14 g. 7) Find the mass of 250.0 mL of benzene. The density of benzene is 0.8786 g/mL. 8) A block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm and weighs 1591.8 g. Calculate the density of lead. 9) 28.5 g of iron shot is added to a graduated cylinder containing 45.5 mL of water. The water level rises to the 49.1 mL mark. Calculate the density of iron. 10) What volume of silver metal will weigh exactly 2500.0 g? The density of silver is 10.5 g/cm³.
The Correct Answer and Explanation is:
Density Formula:Density (p)=Mass (m)Volume (V)\text{Density} \ (p) = \frac{\text{Mass} \ (m)}{\text{Volume} \ (V)}Density (p)=Volume (V)Mass (m)
1)
Given:
Mass = 40.5 g
Volume = 15.0 mL
Solution:p=40.515.0=2.70 g/mLp = \frac{40.5}{15.0} = 2.70 \ \text{g/mL}p=15.040.5=2.70 g/mL
2)
Given:
Mass = 306.0 g
Volume = 22.5 mL
Solution:p=306.022.5=13.6 g/mLp = \frac{306.0}{22.5} = 13.6 \ \text{g/mL}p=22.5306.0=13.6 g/mL
3)
Given:
Density = 0.789 g/mL
Volume = 200.0 mL
Solution (rearranged):Mass=Density×Volume=0.789×200.0=157.8 g\text{Mass} = \text{Density} \times \text{Volume} = 0.789 \times 200.0 = 157.8 \ \text{g}Mass=Density×Volume=0.789×200.0=157.8 g
4)
Given:
Mass = 1896 g
Dimensions: 8.4 cm × 5.5 cm × 4.6 cmVolume=8.4×5.5×4.6=212.52 cm3\text{Volume} = 8.4 \times 5.5 \times 4.6 = 212.52 \ \text{cm}^3 Volume=8.4×5.5×4.6=212.52 cm3p=1896212.52≈8.92 g/cm3p = \frac{1896}{212.52} \approx 8.92 \ \text{g/cm}^3p=212.521896≈8.92 g/cm3
5)
Given:
Mass of liquid = 703.55 g – 345.8 g = 357.75 g
Volume = 225 mL
Solution:p=357.75225=1.59 g/mLp = \frac{357.75}{225} = 1.59 \ \text{g/mL}p=225357.75=1.59 g/mL
6)
Given:
Mass = 65.14 g
Volume = 35.4 mL
Solution:p=65.1435.4≈1.84 g/mLp = \frac{65.14}{35.4} \approx 1.84 \ \text{g/mL}p=35.465.14≈1.84 g/mL
7)
Given:
Density = 0.8786 g/mL
Volume = 250.0 mL
Solution:Mass=0.8786×250.0=219.65 g\text{Mass} = 0.8786 \times 250.0 = 219.65 \ \text{g}Mass=0.8786×250.0=219.65 g
8)
Given:
Mass = 1591.8 g
Dimensions = 4.50 × 5.20 × 6.00 = 140.4 cm³
Solution:p=1591.8140.4≈11.34 g/cm3p = \frac{1591.8}{140.4} \approx 11.34 \ \text{g/cm}^3p=140.41591.8≈11.34 g/cm3
9)
Given:
Mass = 28.5 g
Volume change = 49.1 mL – 45.5 mL = 3.6 mL
Solution:p=28.53.6≈7.92 g/mLp = \frac{28.5}{3.6} \approx 7.92 \ \text{g/mL}p=3.628.5≈7.92 g/mL
10)
Given:
Mass = 2500.0 g
Density = 10.5 g/cm³
Solution:Volume=2500.010.5≈238.1 cm3\text{Volume} = \frac{2500.0}{10.5} \approx 238.1 \ \text{cm}^3Volume=10.52500.0≈238.1 cm3
Explanation
Understanding density helps explain how heavy a substance is for a specific volume. The formula used is density equals mass divided by volume. This means if you know how much something weighs and how much space it takes up, you can figure out how tightly packed the matter inside it is.
Problems 1, 2, 4, 5, 6, 8, and 9 use the direct application of the formula by dividing mass by volume. Others, such as problems 3 and 7, involve finding mass by multiplying density by volume. Problem 10 reverses the process to solve for volume.
Volume is measured in milliliters or cubic centimeters, and mass is usually in grams. Because 1 mL equals 1 cm³ for liquids and solids, the units can be used interchangeably in density problems.
For solids with known dimensions, like in problems 4 and 8, you calculate volume by multiplying length, width, and height. If a liquid fills a container, its volume is the amount the container holds. In displacement methods, like problem 9, the volume of the object is found by how much water it pushes out of the way.
By carefully identifying what’s given and what’s being asked, you can use the formula correctly. Always check your units and make sure you understand which variable you are solving for—density, mass, or volume.
