Question

An element forms a body centered cubic (bcc) lattice with edge length of $300pm$. If the density of the element is $7.2gc{m}^{-3}$, the number of atoms present in $324g$ of it approximately is

• A. $3.33\times 10^{23}$
• B. $6.66\times 10^{23}$
• C. $3.33\times 10^{24}$
• D. $6.66\times 10^{24}$

Answer 1

Answer: (C)

Volume of bcc unit cell = $300pm$

$={\left(300\times 10^{-10}cm\right)}^3=27\times 10^{-24}c{m}^3$

Volume of $324g$ of the element $=\frac{\text{ Mass }}{\text{ Density }}$

$=\frac{324g}{7.2gc{m}^{-3}}=45c{m}^3$

For a bcc structure, number of atoms per unit cell $=2$

So, number of atoms $=\frac{\text{ Total volume }}{\text{ Volume of a unit cell }}$

$=\frac{2\times 45}{27}\times 10^{24}=3.33\times 10^{24}$