If each edge of a cubic unit cell of an element having atomic mass 120 and density $6.25 \mathrm{~g}…
If each edge of a cubic unit cell of an element having atomic mass 120 and density $6.25 \mathrm{~g} \mathrm{cc}^{-1}$ measures $400 \mathrm{pm}$, then the crystal lattice is
primitive
body centered
face centered
end centered
Solution
Atomic weight $(m)=120$
$N_A=6.022 \times 10^{23}$
Unit cell $(a)=400 \mathrm{pm}$
$=400 \times 10^{-10} \mathrm{~m}$
As we know
$d=\frac{Z M}{N_A \times a^3}$
$6.25=\frac{Z \times 120}{6.022 \times 10^{23} \times\left(400 \times 10^{-10}\right)^3}$
$Z=2$
So, the crystal lattice is body centered.