Problem 152
Question
The density of lead (Pb) is 11.3 \(\mathrm{g} / \mathrm{cm}^{3}\) . Calculate the volume of 1 \(\mathrm{mol}\) of \(\mathrm{Pb}\) .
Step-by-Step Solution
Verified Answer
The volume of 1 mol of lead (Pb) is approximately 18.34 cm³.
1Step 1: Find the molar mass of lead (Pb)
To find the molar mass of lead (Pb), we can refer to the periodic table. The molar mass of Pb is approximately 207.2 g/mol.
2Step 2: Calculate the mass of 1 mol of lead (Pb)
Now that we know the molar mass of Pb, we can calculate the mass of 1 mol of Pb. Since 1 mol has a molar mass of 207.2 g/mol, the mass of 1 mol of Pb is:
Mass = (1 mol) × (207.2 g/mol) = 207.2 g
3Step 3: Find the volume of 1 mol of lead (Pb) using density
We are given the density of lead as 11.3 g/cm³. We can use the formula for density to find the volume:
Density = Mass / Volume
Now, we can rearrange the formula to find the volume:
Volume = Mass / Density
We know the mass of 1 mol of Pb (207.2 g) and the density (11.3 g/cm³). Now we can substitute these values into the equation:
Volume = (207.2 g) / (11.3 g/cm³) = 18.34 cm³
4Step 4: State the final answer
The volume of 1 mol of lead (Pb) is approximately 18.34 cm³.
Key Concepts
Molar Mass of LeadVolume Calculation from DensityLead (Pb) Density
Molar Mass of Lead
The molar mass of an element is a critical concept in chemistry. It represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For lead (Pb), the molar mass is approximately 207.2 g/mol.
This value is derived from the periodic table, where atomic masses are listed for each element. Understanding molar mass allows chemists to convert between the number of moles and the mass of a substance. This conversion is essential for various calculations, such as predicting how much of a substance will react or how much will be produced in a chemical process.
This value is derived from the periodic table, where atomic masses are listed for each element. Understanding molar mass allows chemists to convert between the number of moles and the mass of a substance. This conversion is essential for various calculations, such as predicting how much of a substance will react or how much will be produced in a chemical process.
- Molar mass provides a way to bridge the atomic world and the macroscopic world we observe.
- It allows for precise calculations in stoichiometry, ensuring that reactions are balanced.
Volume Calculation from Density
Volume calculation is a common task in chemistry, often derived from mass and density. The density formula, \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \), helps us find volume when the other two values are known.
To find the volume of 1 mole of lead, we use its known density, 11.3 g/cm³, and its mass, 207.2 g. By rearranging the density formula, we can solve for volume:
To find the volume of 1 mole of lead, we use its known density, 11.3 g/cm³, and its mass, 207.2 g. By rearranging the density formula, we can solve for volume:
- \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \)
- Substitute the values: \( \text{Volume} = \frac{207.2\, \text{g}}{11.3\, \text{g/cm}^3} \)
- This results in a volume of approximately 18.34 cm³.
Lead (Pb) Density
Density is a fundamental property in materials science, indicating how much mass is contained within a specific volume. For lead (Pb), a dense metal, this value is 11.3 g/cm³.
- Density affects how substances behave and interact, influencing everything from buoyancy to packaging.
- Knowing the density of lead is essential for determining how much space a given mass will occupy, which is why it was vital in calculating the volume of 1 mol of lead.
Other exercises in this chapter
Problem 149
Calculate the number of sodium ions present in 25.0 g of sodium chloride.
View solution Problem 150
Determine the number of oxygen atoms present in 25.0 g of carbon dioxide.
View solution Problem 153
Explain what is meant by percent composition
View solution Problem 154
What information must a chemist obtain in order to determine the empirical formula of an unknown compound?
View solution