Question

The atomic radius of a hypothetical face-centered cubic (FCC) metal is \((\sqrt{2}/10)\) nm. The atomic weight of the metal is 24.092 g/mol. Taking Avogadro's number to be \(6.023 \times 10^{23}\) atoms/mol, the density of the metal is ............... kg/m³. (Answer in integer)

Hint:

For crystallography problems, memorize the key parameters for common structures: - {FCC:} \(n=4\), \(a=2\sqrt{2}R\), APF = 0.74 - {BCC:} \(n=2\), \(a=4R/\sqrt{3}\), APF = 0.68 - {HCP:} \(n=6\), \(a=2R\), APF = 0.74 Having these at your fingertips is crucial for speed and accuracy.

Answer 1

Step 1: Understanding the Concept:
The theoretical density of a crystalline material can be calculated from its crystal structure properties. The formula relates the mass of the atoms in a unit cell to the volume of the unit cell.
Step 2: Key Formula or Approach:
The formula for the density (\(\rho\)) is: \[ \rho = \frac{n \cdot A}{V_C \cdot N_A} \] where:
- \(n\) = number of atoms per unit cell. For a Face-Centered Cubic (FCC) structure, \(n=4\).
- \(A\) = atomic weight of the metal.
- \(V_C\) = volume of the unit cell.
- \(N_A\) = Avogadro's number.
For an FCC structure, the lattice parameter (\(a\)) is related to the atomic radius (\(R\)) by \(a = 2\sqrt{2}R\). The volume of the cubic unit cell is \(V_C = a^3\).
Step 3: Detailed Calculation:
1. Given Data:
- Atomic radius, \(R = \frac{\sqrt{2}}{10}\) nm.
- Atomic weight, \(A = 24.092\) g/mol.
- Avogadro's number, \(N_A = 6.023 \times 10^{23}\) atoms/mol.
- Crystal structure is FCC, so \(n=4\).
2. Calculate the lattice parameter (a): \[ a = 2\sqrt{2}R = 2\sqrt{2} \left(\frac{\sqrt{2}}{10}\right) \text{nm} = 2 \times \frac{2}{10} \text{nm} = 0.4 \text{ nm} \] Convert to meters: \(a = 0.4 \times 10^{-9}\) m.
3. Calculate the volume of the unit cell (\(V_C\)): \[ V_C = a^3 = (0.4 \times 10^{-9} \text{ m})^3 = 0.064 \times 10^{-27} \text{ m}^3 \] 4. Calculate the density (\(\rho\)):
Substitute all values into the density formula: \[ \rho = \frac{4 \text{ atoms/cell} \times 24.092 \text{ g/mol}}{(0.064 \times 10^{-27} \text{ m}^3/\text{cell}) \times (6.023 \times 10^{23} \text{ atoms/mol})} \] \[ \rho = \frac{96.368 \text{ g}}{0.064 \times 6.023 \times 10^{-4} \text{ m}^3} \] \[ \rho = \frac{96.368}{0.385472 \times 10^{-4}} \frac{\text{g}}{\text{m}^3} \approx 250.0 \times 10^4 \frac{\text{g}}{\text{m}^3} \] \[ \rho = 2,500,000 \text{ g/m}^3 \] 5. Convert to kg/m³:
Since \(1 \text{ kg} = 1000 \text{ g}\), \[ \rho = \frac{2,500,000}{1000} \frac{\text{kg}}{\text{m}^3} = 2500 \text{ kg/m}^3 \] Step 4: Final Answer:
The density of the metal is 2500 kg/m³.
Step 5: Why This is Correct:
The calculation correctly uses the properties of the FCC crystal structure (\(n=4\), \(a=2\sqrt{2}R\)) and the standard formula for theoretical density. All unit conversions are performed correctly to arrive at the final answer in kg/m³.