Question

Silver crystallises in f.c.c. lattice. If edge length of the cell is $4.077 \times 10^{-8}$ cm and density is $10.5 \, \text{gm cm}^{-3}$, calculate the atomic mass of silver.

Hint:

For f.c.c. lattices, always use $Z = 4$. For b.c.c., $Z = 2$ and for simple cubic, $Z = 1$.

Answer 1

Step 1: Formula for density.
\[ d = \frac{Z \times M}{N_A \times a^3} \] where, \(d\) = density of the crystal = $10.5 \, \text{gm cm}^{-3}$ 
\(Z\) = number of atoms per unit cell = 4 (for f.c.c.) 
\(M\) = molar mass of silver (to be calculated) 
\(N_A\) = Avogadro’s number = $6.022 \times 10^{23} \, mol^{-1}$ 
\(a\) = edge length = $4.077 \times 10^{-8}$ cm 

Step 2: Calculate unit cell volume.
\[ a^3 = (4.077 \times 10^{-8})^3 \, \text{cm}^3 = 6.79 \times 10^{-23} \, \text{cm}^3 \] 

Step 3: Apply values in formula.
\[ 10.5 = \frac{4 \times M}{6.022 \times 10^{23} \times 6.79 \times 10^{-23}} \]

 Step 4: Simplify.
\[ 10.5 = \frac{4M}{40.9} \] \[ M = \frac{10.5 \times 40.9}{4} \approx 107.3 \] 

Conclusion:
The atomic mass of silver is: \[ \boxed{107.3 \, \text{g mol}^{-1}} \]