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.

Explanation

`d=(zm)/(a^(3)N)`, m = Mass of element, N = number of atoms
Or `N=(108xx4)/(10.8xx27xx10^(-24))=1.48xx10^(24)` atoms
`M=(a^(3)xxN_(a)xxd)/(Z)`
`=(27xx10^(-24)xx6.022xx10^(23)xx10.8)/(4)=43.88"g mol"^(-1)`
`43.88"g mol"^(-1)` contains `6.02xx10^(23)` atoms
So, 108 g contains `=(6.02xx10^(23)xx108)/(43.88)=1.48xx10^(24)` atoms