An element has a bcc structure with a cell edge of `288` pm. The density of the element is `7.2 g cm^(-3)`. How many atoms are present in `208 g` of the element?

Explanation

Volume of the unit cell `=(288xx10^(-10))^(3)=23.9xx10^(-24)cm^(3)`
Volume of 208 g of the element `=(208)/(7.2)=28.88cm^(3)`
Number of unit cells in `28.88cm^(3)=(28.88)/(23.9xx10^(-24))=12.08xx10^(23)` unit cells
Each bcc structure contains 2 atoms.
So total atoms in `12.08xx10^(23)"unit cells"=2xx12.08xx10^(23)=24.16xx10^(23)`