Problem

Silver (Ag) crystallises in FCC structure. Given:

• Molar mass M = 108 g/mol
• Edge length a = 4.077 Å
• Nₐ = $6.022 \times 10^{23}$ $mol^{-1}$

Calculate the density of silver.

Solution

Step 1: Identify Z FCC structure → Z = 4 atoms per unit cell.

Step 2: Convert edge length to cm $a = 4.077 \text{ Å} = 4.077 \times 10^{-8} \text{ cm}$

Step 3: Calculate $a^{3}$ $a^3 = (4.077 \times 10^{-8})^3 \text{ cm}^3$ $= (4.077)^3 \times 10^{-24} \text{ cm}^3$ $= 67.76 \times 10^{-24} \text{ cm}^3$ $= 6.776 \times 10^{-23} \text{ cm}^3$

Step 4: Apply density formula $\rho = \frac{Z \cdot M}{a^3 \cdot N_A} = \frac{4 \times 108}{6.776 \times 10^{-23} \times 6.022 \times 10^{23}}$

Step 5: Calculate denominator $6.776 \times 6.022 = 40.79$

Step 6: Final answer $\rho = \frac{432}{40.79} \approx \boxed{10.59 \text{ g/cm}^3}$

Verification: Actual density of silver = 10.5 g/$cm^{3}$. Close agreement confirms the calculation.

Common errors to avoid:

1. Using Z = 2 (BCC) instead of Z = 4 (FCC)
2. Forgetting to cube the edge length
3. Not converting Å to cm (forgetting × 10^{-8})