Question

Molybdenum has a atomic mass $96gmo{l}^{-1}$ with density $10.3gc{m}^{-3}$.The edge length of unit cell is $314$pm.Determine its lattice structure.

Answer 1

Density of element, $\rho$ = 10.3 g ${cm}^{-3}$

Cell edge a=314 pm

$N_A=6.023\times {10}^{23}{mol}^{-1}$

Atomic mass = 96 g ${mol}^{-1}$

$P=\frac{Z\times M}{a^3\times N_A}$

$Z=\frac{P\times N_A\times a^3}{M}$

$=\frac{10g{cm}^{-3}\times {10}^{-30}{cm}^3\times 6.023\times {10}^{23}{mol}^{-1}}{96g{mol}^{-1}}$ = 2

The structure of the crystal lattic is B.C.C