An element crystallizes in fec lattice with a cell edge of 300 pm. The density of the element is 10.8 g `cm^(-3)`. Calculate the number of atoms in 108 g of the element.

Answer

`1.48 xx 10^(24)`

Explanation

Volume of 108 g element = `108/10.8 = 10cm^3`
volume of unit cell = `(300 xx 10^(-10) cm)^(3) = 27 xx 10^(-24) cm^3`
No. of unit cells = `(10)/(27 xx 10^(-24)) = 3.7 xx 10^(23) ` unit cells
Since the structure is fcc, number of atoms per unit cell = 4
No. of atoms in `108 g = 3.7 xx 10^(23) xx 4`
`= 1.48 xx 10^(24)`