Pd has an atomic radius of 0.137 nm, an atomic weight of 106.4 g/mol, and it forms a face-centered cubic structure. Estimate the density of pure Pd. Give your answer in g/cm^3.

To estimate the density of pure Palladium (Pd), we can follow these steps:

### 1. Atomic mass and atomic radius:
- Atomic radius (\(r\)) = 0.137 nm = \(1.37 \times 10^{-8}\) cm
- Atomic weight (M) = 106.4 g/mol

### 2. Palladium has a face-centered cubic (FCC) structure:
In an FCC structure, the relationship between the atomic radius (\(r\)) and the lattice parameter (\(a\)) is given by:

\[
a = 2\sqrt{2}r
\]

Now, substitute the atomic radius \(r = 1.37 \times 10^{-8}\) cm:

\[
a = 2\sqrt{2} \times 1.37 \times 10^{-8} \, \text{cm}
\]
\[
a = 3.874 \times 10^{-8} \, \text{cm}
\]

### 3. Volume of the unit cell:
The volume of a cube with side length \(a\) is:

\[
V_{\text{cell}} = a^3
\]

\[
V_{\text{cell}} = (3.874 \times 10^{-8})^3 \, \text{cm}^3
\]
\[
V_{\text{cell}} = 5.81 \times 10^{-23} \, \text{cm}^3
\]

### 4. Number of atoms per unit cell:
For a face-centered cubic (FCC) structure, there are 4 atoms per unit cell.

### 5. Moles of atoms in one unit cell:
The number of atoms per unit cell is 4. To convert this into moles, we use Avogadro's number (\(N_A = 6.022 \times 10^{23}\) atoms/mol):

\[
\text{moles of atoms per unit cell} = \frac{4}{6.022 \times 10^{23}} = 6.64 \times 10^{-24} \, \text{mol}
\]

### 6. Mass of atoms in one unit cell:
Now, calculate the mass of the atoms in one unit cell using the atomic weight (M):

\[
\text{mass of unit cell} = 6.64 \times 10^{-24} \, \text{mol} \times 106.4 \, \text{g/mol}
\]
\[
\text{mass of unit cell} = 7.06 \times 10^{-22} \, \text{g}
\]

### 7. Density:
Finally, the density (\(\rho\)) is mass divided by volume:

\[
\rho = \frac{\text{mass of unit cell}}{V_{\text{cell}}}
\]
\[
\rho = \frac{7.06 \times 10^{-22} \, \text{g}}{5.81 \times 10^{-23} \, \text{cm}^3}
\]
\[
\rho \approx 12.15 \, \text{g/cm}^3
\]

### Final Answer:
The density of pure Palladium (Pd) is approximately **12.15 g/cm³**.

Would you like more details or have any questions?

Here are five related questions:
1. What is the relationship between the atomic radius and lattice parameter in other crystal structures?
2. How do you calculate the number of atoms in different unit cell structures like body-centered cubic (BCC)?
3. How does the density of Palladium compare to other transition metals?
4. What are the factors that can affect the density of materials in real-world applications?
5. Can you calculate the volume of a single Palladium atom based on its atomic radius?

**Tip:** The structure of the material plays a significant role in determining its density, as the arrangement of atoms affects the volume they occupy.

Learn how to estimate the density of pure Palladium (Pd) with a face-centered cubic (FCC) structure. This solution covers the calculation of the lattice parameter, volume of the unit cell, number of atoms, and density, using an atomic radius of 0.137 nm and atomic weight of 106.4 g/mol. Suitable for students in Grades 11-12.

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