Sodium crystallises in a bcc lattice having unit cell with an edge length of 4.29Å. Calculate the number of unit cells present in 2.3g of Na metal. Determine the number of atoms present in 1 cc of unit cell.

Explanation

The number of Na atoms in 2.3g of Na metal
`=2.3/23xx6.022xx10^(23)=6.022xx10^(22)`
As Na crystallises in a bcc lattice, its unit cell contains 2 Na atoms.
Therefore, the number of unit cells in 2.3g of Na metal
`=1/2xx6.022xx10^(22)=3.0115xx10^(22)`
The volume of 1 unit cell
`=(4.29xx10^(-8))^(3)cm^(3)=7.8953xx10^(-23)cm^(3)`
As 1 unit cell contains 2 Na atoms, `7.8953xx10^(-23)cm^(3)` of volume is associated with 2 Na atoms. This means that the number of Na atoms per `cm^(3)` is
`2/(7.8953xx10^(-23))=2.533xx10^(22)`