Q.Silver crystallises in fcc lattice. If edge length of the cell is $4.07\times10^{-8}$ cm and density is 10.5 g cm$^{-3}$, calculate the atomic mass of silver.

$a^3 = (4.07\times10^{-8})^3 = 6.74\times10^{-23}$ cm$^3$. …

Step 1 – Formula.

Step 2 – Cell volume.

Step 3 – Substitute.

$M = \frac{10.5\times 6.74\times10^{-23}\times 6.022\times10^{23}}{4}$ …

Rearrange $d=zM/(a^3N_A)$ to $M=d a^3 N_A/z$, with $z=4$ for fcc; c …

• Using $z=2$ (bcc) instead of $4$ for fcc.
• Not cubing the edge length or a power-of-ten slip. …