Question

Silver forms CCP structure and its density is $10.5~\text{g/cm}^3$. What is the edge length of the unit ll? (Molar mass of silver is $107.9~\text{g/mol}$)

• A. $\sqrt[3]{0.68}~\text{Å}$
• B. $\sqrt[3]{48}~\text{Å}$
• C. $\sqrt[3]{68.1}~\text{Å}$
• D. $\sqrt[3]{680}~\text{Å}$

Hint:

CCP Lattice and Density Relation.

• Use: $\rho = \fracZ Ma^3 N_A$
• $Z = 4$ for FCC (ccp).
• Ensure units of mass (g), volume (cm$^3$), and $N_A$ match.
• Convert final $a$ to Å ($1~\textcm = 10^8~\textÅ$).

Answer 1

The Correct Option is C

For a cubic close-packed (ccp/fcc) structure, the number of atoms per unit cell = 4.
Density formula: \[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \] Given: \[ \rho = 10.5~\text{g/cm}^3,~M = 107.9~\text{g/mol},~Z = 4,~N_A = 6.022 \times 10^{23} \] Solving for $a$: \[ a^3 = \frac{Z \cdot M}{\rho \cdot N_A} = \frac{4 \cdot 107.9}{10.5 \cdot 6.022 \times 10^{23}}~\text{cm}^3 \] Convert to Å using $1~\text{cm} = 10^{8}~\text{Å}$: \[ a \approx \sqrt[3]{68.1}~\text{Å} \]

Answer 2

Step 1: Understanding the problem
Silver forms a cubic close-packed (CCP) structure, which is the same as face-centered cubic (FCC). We are asked to find the edge length (a) of the unit cell.
Given data:
- Density (ρ) = 10.5 g/cm³
- Molar mass (M) = 107.9 g/mol
- Structure type = CCP (FCC)

Step 2: Number of atoms per unit cell
In CCP (FCC), there are 4 atoms per unit cell.

Step 3: Calculate the mass of one unit cell
Mass of one mole = 107.9 g contains Avogadro’s number (N_A = 6.022 × 10²³) of atoms.
Mass of one atom = 107.9 g / (6.022 × 10²³) = 1.792 × 10⁻²² g
Mass of unit cell = 4 atoms × 1.792 × 10⁻²² g = 7.168 × 10⁻²² g

Step 4: Calculate volume of the unit cell using density
Density = Mass / Volume ⇒ Volume = Mass / Density
Volume of unit cell = (7.168 × 10⁻²² g) / (10.5 g/cm³) = 6.83 × 10⁻²³ cm³

Step 5: Calculate edge length (a)
Since the unit cell is cubic, volume = a³
\[ a = \sqrt[3]{6.83 \times 10^{-23}} \text{ cm} = 4.08 \times 10^{-8} \text{ cm} = 4.08 \text{ Å} \]

Step 6: Verification with given answer
The problem’s answer form is \(\sqrt[3]{68.1} \, \text{Å}\).
Note: \(68.1 \times 10^{-24} \text{ cm}^3 = 6.81 \times 10^{-23} \text{ cm}^3\), close to our calculated volume.
Taking cube root of 68.1 ų also gives approx 4.08 Å.

Step 7: Conclusion
Therefore, the edge length of the silver unit cell is \(\sqrt[3]{68.1} \, \text{Å}\).