Question

An element crystallises in a bcc lattice with cell edge of $500$ pm. The density of the element is $7.5$g $c{m}^{-3}$. How many atoms are present in $300$ g of the element?

Answer 1

The edge length, $a=500pm=500\times {10}^{-12}m=5\times {10}^{-8}cm$
The volume of the unit cell $=a^3=(5\times {10}^{-8}cm{)}^3=1.25\times {10}^{-22}c{m}^3$
For b.c.c unit cell, the number of atoms per unit cell $Z=2$
Volume occupied by each atom $=\frac{1.25\times {10}^{-22}c{m}^3}{2}=6.25\times {10}^{-23}c{m}^3$
The density $d=7.5g/c{m}^3$
Mass of sample $=300g$
Volume of sample $=\frac{mass}{density}=\frac{300}{7.5}=40c{m}^3$
The number of atoms present in 300 g of the element $=\frac{40c{m}^3}{6.25\times {10}^{-23}c{m}^3}=6.4\times {10}^{23}$ atoms.