Question

Sliver forms a ccp lattice. The edge length of its unit cell is 408.6 pm. Calculate the density of sliver?$(N_A=6.022\times {10}^{23}$, Atomic mass of $Ag=108gmo{l}^{-1})$

Answer 1

The edge length of its unit cell is 408.6 pm or
$408.6\times {10}^{-10}$ cm
The volume of the unit cell $=a^3=(408.6\times {10}^{-10}cm{)}^3=68.27\times {10}^{-24}c{m}^3$
ccp unit cell has 4 atoms per unit cell.
Mass of one $Ag$ atom
$=\frac{\text{molar mass of Ag}}{\text{Avogadro's number}}=\frac{108g/mol}{6.022\times {10}^{23}/mol}=17.93\times {10}^{-23}g$
Mass of unit cell $=4\times 17.93\times {10}^{-23}g=71.72\times {10}^{-23}g$
$\text{Density of silver}=\frac{\text{mass of unit cell}}{\text{volume of unit cell}}$
$\text{Density of silver}=\frac{71.72\times {10}^{-23}g}{68.27\times {10}^{-24}c{m}^3}=10.51g/c{m}^3$