Question

Calculate the density of silver which crystallizes in face centred form. The distance between nearest metal atoms is 287 pm. $(Ag=107.87g{mol}^{-1}$, $N_A=6.022{\times 10}^{23}$ atoms ${mol}^{-1}$)

Answer 1

Theoretical density of crystal, $\rho =\frac{nM}{N_A{a}^3}g/c{m}^3$

For FCC, $n=4$ and distance between nearest atoms in FCC is given by $d=\frac{a}{\sqrt{2}}$ where a is the edge length.

$\Rightarrow 287\times {10}^{-10}cm=\frac{a}{\sqrt{2}}$

$\Rightarrow a=404\times 67\times {10}^{-10}cm$

So, $\rho =\frac{4\times 107.87}{6.022\times {10}^{23}\times (404.67{)}^3\times {10}^{-30}}=10.81g/c{m}^3$