Question

An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is $10.8gc{m}^{-3}$. Calculate the number of atoms in 108 g of the element.

Answer 1

volume of 108 g of elements$=\frac{mass}{density}=10c{m}^3$
Volume of one unit cell $=(300pm{)}^3=2.7\times {10}^{-23}c{m}^3$
Numbers of unit cell in $10c{m}^3=\frac{10}{2.7\times {10}^{-23}c{m}^3}=\frac{{10}^{24}}{2.7}unitcells$
one unit cell have 4 atoms so total numbers of atoms $=\frac{{10}^{24}}{2.7}unitcells\times 4=1.48\times {10}^{24}atoms$