Question

Molybdenum forms body-centred cubic crystals whose density is $10.3gc{m}^{-3}$$.Calculatetheedgelengthoftheunitcell.Themolarmassof$Mo$is$95.94\ g\ mol^{-1}$.

Answer 1

$\rho =\frac{Z\times M_A}{N_A\times a^3}$

$Z=2$ (for BCC) , $\rho$=$10.3g/{cm}^3$ $M_A=95.94$ $gmo{l}^{-1}$

$10.3=\frac{2\times 95.94}{6.023\times {10}^{23}\times a^3}$

$\Rightarrow a^3=31.14\times {10}^{-24}$ $c{m}^3$

$\Rightarrow a=3.146\mathring{A}$.