Problem 155
Question
The density of gold is \(19.3 \mathrm{~g} / \mathrm{mL}\), that of lead is \(11.4 \mathrm{~g} / \mathrm{mL}\), that of iron \(7.8 \mathrm{~g} / \mathrm{mL}\), and that of aluminum \(2.7 \mathrm{~g} / \mathrm{mL}\). A student is given separate samples of three substances \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), along with a graduated cylinder containing \(50.0 \mathrm{~mL}\) of water. Each sample has a mass of \(200.0 \mathrm{~g}\), and the student finds that the volumes of the samples are A, \(25.64 \mathrm{~mL} ; \mathrm{B}, 10.36 \mathrm{~mL} ;\) and \(\mathrm{C}, 17.54 \mathrm{~mL}\). What is the identity of each substance?
Step-by-Step Solution
Verified Answer
The identities of the substances are: Substance A is Iron, Substance B is Gold, and Substance C is Lead.
1Step 1: Calculate the density of substance A
To calculate the density of substance A, we will use the given mass (200.0 g) and volume (25.64 mL) and the formula mentioned above:
Density of A = Mass of A / Volume of A = \( \frac{200.0 \mathrm{~g}}{25.64 \mathrm{~mL}} = 7.81 \mathrm{~g/mL} \)
2Step 2: Calculate the density of substance B
Similarly, using the mass (200.0 g) and volume (10.36 mL) for substance B, we will calculate its density:
Density of B = Mass of B / Volume of B = \( \frac{200.0 \mathrm{~g}}{10.36 \mathrm{~mL}} = 19.30 \mathrm{~g/mL} \)
3Step 3: Calculate the density of substance C
Finally, using the mass (200.0 g) and volume (17.54 mL) for substance C, we will calculate its density:
Density of C = Mass of C / Volume of C = \( \frac{200.0 \mathrm{~g}}{17.54 \mathrm{~mL}} = 11.40 \mathrm{~g/mL} \)
4Step 4: Identify the substances
Now that we have the calculated densities for substances A, B, and C, we can compare them to the given densities of gold, lead, iron, and aluminum:
- Density of substance A = 7.81 g/mL ≈ 7.8 g/mL, that of iron. Therefore, substance A is iron.
- Density of substance B = 19.30 g/mL ≈ 19.3 g/mL, that of gold. Therefore, substance B is gold.
- Density of substance C = 11.40 g/mL ≈ 11.4 g/mL, that of lead. Therefore, substance C is lead.
So, the final identification of the substances is as follows:
- Substance A: Iron
- Substance B: Gold
- Substance C: Lead
Key Concepts
Density of MetalsMass and Volume RelationshipIdentifying Substances by Density
Density of Metals
Density is the measure of how much mass is contained in a given volume. The formula for density is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \), typically expressed in grams per milliliter (g/mL) or kilograms per cubic meter (kg/m³). For metals, density can give us a lot of information about the material since different metals have characteristic densities.
- Gold has a density of 19.3 g/mL, making it one of the heaviest and densest metals.
- Lead, with a density of 11.4 g/mL, is also quite dense but less so than gold.
- Iron has a density of 7.8 g/mL, which is moderate compared to other metals.
- Aluminum is relatively light with a density of 2.7 g/mL.
Mass and Volume Relationship
The relationship between mass and volume is crucial in understanding density. Mass is the amount of substance in an object, measured in grams or kilograms, while volume is the amount of space it occupies, measured in milliliters (mL) or liters (L).
When given a mass and you need the volume, you can rearrange the density formula: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). Similarly, to calculate mass if you know the density and volume, use: \( \text{Mass} = \text{Density} \times \text{Volume} \).
This basic calculation is used in exercises to determine how a particular material compares to its expected density values. For the exercise, all samples had fixed mass (200.0 g) but their volumes varied:
When given a mass and you need the volume, you can rearrange the density formula: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). Similarly, to calculate mass if you know the density and volume, use: \( \text{Mass} = \text{Density} \times \text{Volume} \).
This basic calculation is used in exercises to determine how a particular material compares to its expected density values. For the exercise, all samples had fixed mass (200.0 g) but their volumes varied:
- Sample A had a volume of 25.64 mL.
- Sample B's volume was 10.36 mL.
- Sample C measured 17.54 mL.
Identifying Substances by Density
Identifying substances using density is a practical and straightforward method. Each substance has a unique density, acting as a fingerprint. By measuring a sample's mass and volume, and subsequently calculating its density, you can compare it to known densities of other substances to identify it.
In the given exercise, the calculated densities were:
In the given exercise, the calculated densities were:
- Sample A: 7.81 g/mL
- Sample B: 19.30 g/mL
- Sample C: 11.40 g/mL
- Sample A's density was close to that of iron (7.8 g/mL), so Sample A is identified as iron.
- Sample B's density matched perfectly with gold (19.3 g/mL), thus it is gold.
- Sample C's density aligned with lead (11.4 g/mL), confirming it as lead.
Other exercises in this chapter
Problem 153
Dieters are often told that drinking ice-cold water burns more energy than drinking roomtemperature water. Why is this true?
View solution Problem 154
If a graduated cylinder has markings that indicate \(0.01 \mathrm{~mL}\), what is the uncertainty in any volumes determined using this graduated cylinder?
View solution Problem 156
When \(10 \mathrm{~kJ}\) of heat energy is added to a beaker containing \(250 \mathrm{~g}\) of water initially at \(23.0{ }^{\circ} \mathrm{C}\), what is the fi
View solution Problem 157
Indicate the uncertainty in: (a) \(74.8 \mathrm{~m}\) (b) \(0.0026 \mathrm{~g}\) (c) \(1.250 \times 10^{3} \mathrm{~L}\) (d) \(18 \mathrm{~cm}\) (e) 18 pennies
View solution