Question

Silver crystallises in fcc lattice. If edge length of the cell is $4.07\times {10}^{-8}cm$ and density is $10.5gc{m}^{-3}$, calculate the atomic mass of silver.

Answer 1

Density $\left(D\right)$, number of atoms per unit cell $\left(Z\right)$, atomic mass $\left(M\right)$, Avogadro's number $\left(N_0\right)$ and edge length ($a)$ are related as below:

$D=\frac{ZM}{N_0{a}^3}$
$10.5=\frac{4M}{6.023\times {10}^{23}\times (4.07\times {10}^{-8}{)}^3}$

$M=107.9$ g/mol

Hence, the atomic mass of silver is $107.9$ g/mol.