Question

An element crystallizes in a FCC lattice with cell edge of 400 pm. The density of the element is $7gc{m}^{-3}$. How many atoms are present in 280 g of the element?

Answer 1

Volume of the unit cell = $(400pm{)}^3$
$=6.4\times {10}^{7}p{m}^3$
$=6.4\times {10}^{-23}c{m}^3$
Volume of 280 g of the element $=\frac{280}{7}=40c{m}^3$
No. of unit cell in this volume $=\frac{40c{m}^3}{6.2\times {10}^{-23}c{m}^3/\text{unit cell}}$
$=6.4\times {10}^{23}\text{unit cell}$
Since, each FCC cubic unit cell contains 4 atoms.