Question

Silver forms ccp lattice. Edge length of its unit cell is 408.6 pm. Calculate the density of silver. (Atomic weight of Ag = 108)

Hint:

For ccp/fcc structures, always take \(Z=4\).

Answer 1

Step 1: Recall formula for density.
\[ \rho = \frac{Z \times M}{N_A \times a^3} \] where \(Z = 4\) (ccp), \(M = 108 \ g \ mol^{-1}\), \(a = 408.6 \ \text{pm}\), \(N_A = 6.022 \times 10^{23}\). Step 2: Convert edge length.
\[ a = 408.6 \times 10^{-10} \ \text{cm} = 4.086 \times 10^{-8} \ \text{cm} \] Step 3: Substitution.
\[ \rho = \frac{4 \times 108}{6.022 \times 10^{23} \times (4.086 \times 10^{-8})^3} \] \[ \rho \approx 10.5 \ g \ cm^{-3} \] Step 4: Conclusion.
Density of silver = \(10.5 \ g \ cm^{-3}\).