Question

Iron has a body centred cubic unit cell with the cell dimension of 286.65 pm. Density of iron is 7.87 g cm${}^{-3}$. Use this information to calculate Avogadro's number?

(Atomic mass of Fe = 56.0 u)

Answer 1

$a$ $=$ $286.65$ pm

$=286.65\times {10}^{-10}cm$

Density $\left(ϵ\right)=7.874gc{m}^{-3}$

At mass of $Fe=$ $56.0$u

$Z=2$(for body centered cubic unit cell)

Avogadro number $\left(N_0\right)=?$

$p=\frac{Z\times M}{a^3\times N_0}$

$7.874gc{m}^{-3}=\frac{2\times \left(56.0gmo{l}^{-1}\right)}{(286.65\times {10}^{-10}cm{)}^3\times N_0}$

$N_0=\frac{2\times \left(56gmo{l}^{-1}\right)}{(286.65\times {10}^{-10}c{m}^3)\times \left(7.874gc{m}^{-3}\right)}$

$=6.022\times {10}^{23}mo{l}^{-1}$