An element 'X' (At. Mass = 40 g `mol^(-1)`) having f.c.c structure has unit cell edge length of 400 pm . Calculate the density of 'X' and the number of unit cells in 4 g of 'X' .
`(N_(A) = 6.022 xx 10^(23) mol^(-1))`.

Explanation

`d=(zM)/(a^3N_A)`
=`(4 times 40)/((4 times 10^-8) times 6.022 times 10^-23)`
`=4.15g//cm^3`
No of unit cells=total no of atoms 4
=`[4/40 times 6.022 times 10^23]//4`
`=1.5 times 10^22`