How many atoms of niobium are present in $2.43 \mathrm{~g}$ if it forms bcc structure with density $9…
How many atoms of niobium are present in $2.43 \mathrm{~g}$ if it forms bcc structure with density $9 \mathrm{~g} \mathrm{~cm}^{-3}$ and volume of unit cell $2.7 \times 10^{-23} \mathrm{~cm}^3 ?$
$3.01 \times 10^{23}$
$4.1 \times 10^{22}$
$5.0 \times 10^{22}$
$2.0 \times 10^{22}$
Solution
Let number of atoms are $\mathrm{N}$
Mass of one atom $=2.43 / \mathrm{N}$
$\mathrm{d}=\frac{\mathrm{Z} \times \text { mass of one atom }}{\text { Volume of unit cell }}[\mathrm{Z}=2$ for $\mathrm{BCC}]$
$9=\frac{2 \times 2.43 / \mathrm{N}}{2.7 \times 10^{-23}} \quad \mathrm{~N}=\frac{2 \times 2.43}{2.7 \times 10^{-23} \times 9}=2 \times 10^{22}$