Given –

$\rho =9.6 \times 10^{2} kg/m^{3}$

$a=4.3A^{\circ} = 4.3 \times 10^{-10}m$

$M=23$

To find – $n$, (identify the crystal) Solution – According to, $n=\dfrac {\rho \cdot a^{3}}{\left(\frac{M}{N_A}\right)}$ $n=\dfrac {9.6 \times 10^{2}\times (4.3\times 10^{-10})^{3}}{\frac{23}{6.0238 \times 10^{23}}}=2$ $\because n=2$

∴ The crystal structure is BCC.