An element crystallises in a body-centred cubic crystal structure with the unit cell edge length of 291pm. The density of the element is `7.54g*cm^(-3)`. How many unit cells are there in 168g of the element?

Explanation

We know, `rho=(ZxxM)/(Nxxa^(3))`
Given: Z = 2 (since the element has bcc unit cell), a = 291pm = `291xx10^(-10)cm and rho=7.54g*cm^(-3)`
`therefore" "M=(rhoxxNxxa^(3))/Z=(7.54xx6.022xx10^(23)xx(291xx10^(-10))^(3))/2`
`=55.95g*mol^(-1)`
Therefore, number of atoms in 168g of the element
`=168/55.95xx6.022xx10^(23)=18.08xx10^(23)`
Number of atoms per unit cell of the element = 2
`therefore" "` Number of unit cell in 168g of the element = `1/2xx18.8xx10^(23)=9.04xx10^(23)`