An element crystallises in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 `g*cm^(-3)`. How many atoms are present is 300 g/mol of the element?

Explanation

First we require to known the molar mass (M) of the element. This can be calculated by using the formula-
`M=(rhoxxNxxa^(3))/Z`
For a bcc lattice Z = 2. It is given that, `rho=7.5g*cm^(-3)` and a = 500pm = `5xx10^(-8)cm`
`thereforeM=(7.5xx6.022xx10^(23)xx(5xx10^(-8))^(3))/2g*mol^(-1)`
`=282.28g*mol^(-1)`
So, number of atoms in 300 g of the element
`=300/282.28xx6.022xx10^(23)=6.4xx10^(23)`