A hypothetical metal, which is a single crystal and made up of one chemical element, has an FCC crystal structure. Given the density of the metal is 10 g/cm³ and the atomic weight/mass of the metal is 100 g/mol, estimate the atomic radius of the metal. Note: Avogadro's number Na = 6.022×10²3 atoms/mol %3D
A hypothetical metal, which is a single crystal and made up of one chemical element, has an FCC crystal structure. Given the density of the metal is 10 g/cm³ and the atomic weight/mass of the metal is 100 g/mol, estimate the atomic radius of the metal. Note: Avogadro's number Na = 6.022×10²3 atoms/mol %3D
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![### Estimating the Atomic Radius of a Hypothetical Metal
A hypothetical metal, which consists of a single crystal and is composed of one type of chemical element, has an FCC (Face-Centered Cubic) crystal structure. Given the following data:
- **Density of the metal**: \(10 \, \text{g/cm}^3\)
- **Atomic weight/mass of the metal**: \(100 \, \text{g/mol}\)
**Objective**: Estimate the atomic radius of the metal.
**Note**: Avogadro’s number \(N_A\) is \(6.022 \times 10^{23} \, \text{atoms/mol}\).
#### Explanation of FCC Crystal Structure
In an FCC crystal structure:
- Atoms are located at each of the corners and the centers of all the cube faces of the unit cell.
- There are 4 atoms per unit cell (8 corners * 1/8 atom per corner + 6 faces * 1/2 atom per face).
#### Calculating Molar Volume:
1. **Density Equation**:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Hence, the volume occupied by one mole of the hypothetical metal can be calculated by:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \, \text{g}}{10 \, \text{g/cm}^3} = 10 \, \text{cm}^3 \]
2. **Volume of Metal Per Atom**:
\[ \text{Volume per atom} = \frac{\text{Volume per mole}}{N_A} = \frac{10 \, \text{cm}^3}{6.022 \times 10^{23} \, \text{atoms/mol}} = 1.66 \times 10^{-23} \, \text{cm}^3/\text{atom} \]
3. **Relation between Volume and Atomic Radius in FCC Structure**:
The volume of the unit cell \( V_{\text{cell}} \) can also be related to the atomic radius \( r \) by:
\[
a^3 = \text{Volume of the unit cell}, \quad \text{where} \ a = 2\sqrt{2}r
\]
Since there are 4 atoms](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd9f391b6-5f04-4d98-9bc5-d57c88046352%2F77eb761a-93b4-494e-9f42-0f978083eba2%2Ftx8qlfh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Estimating the Atomic Radius of a Hypothetical Metal
A hypothetical metal, which consists of a single crystal and is composed of one type of chemical element, has an FCC (Face-Centered Cubic) crystal structure. Given the following data:
- **Density of the metal**: \(10 \, \text{g/cm}^3\)
- **Atomic weight/mass of the metal**: \(100 \, \text{g/mol}\)
**Objective**: Estimate the atomic radius of the metal.
**Note**: Avogadro’s number \(N_A\) is \(6.022 \times 10^{23} \, \text{atoms/mol}\).
#### Explanation of FCC Crystal Structure
In an FCC crystal structure:
- Atoms are located at each of the corners and the centers of all the cube faces of the unit cell.
- There are 4 atoms per unit cell (8 corners * 1/8 atom per corner + 6 faces * 1/2 atom per face).
#### Calculating Molar Volume:
1. **Density Equation**:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Hence, the volume occupied by one mole of the hypothetical metal can be calculated by:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \, \text{g}}{10 \, \text{g/cm}^3} = 10 \, \text{cm}^3 \]
2. **Volume of Metal Per Atom**:
\[ \text{Volume per atom} = \frac{\text{Volume per mole}}{N_A} = \frac{10 \, \text{cm}^3}{6.022 \times 10^{23} \, \text{atoms/mol}} = 1.66 \times 10^{-23} \, \text{cm}^3/\text{atom} \]
3. **Relation between Volume and Atomic Radius in FCC Structure**:
The volume of the unit cell \( V_{\text{cell}} \) can also be related to the atomic radius \( r \) by:
\[
a^3 = \text{Volume of the unit cell}, \quad \text{where} \ a = 2\sqrt{2}r
\]
Since there are 4 atoms
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