Question

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

Answer 1

Answer: 22

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