Question

An element M crystallises in a body centred cubic unit cell with a cell edge of $300\, pm$. The density of the element is $6.0\, g \, cm ^{-3}$. The number of atoms present in $180\, g$ of the element is ___$\times 10^{23}$. (Nearest integer)

Answer 1

$M$ is body certred cubic,
$\therefore Z =2$
Let mass of 1 atom of $M$ is $A$
Edge length $=300\, pm$
Density $=6\, g / cm ^3$
$\therefore 6 \, g / cm ^3=\frac{ Z \times A }{\left(300 \times 10^{-10}\right)^3}=\frac{2 \times A }{27 \times 10^{-24}}$
$A =81 \times 10^{-24} g$
$\therefore$ Atomic mass $=48.6\, g$
$\therefore$ Mole in $180 g =\frac{180}{48.6}=3.7\, moles$
Atoms of $M =3.7 \times 6 \times 10^{23}$
$=22.22 \times 10^{23}$ atoms