An element 'X' (atomic mass = 40`g*mol^(-1)`) having fcc structure, has unit cell edge length of 400 pm. Calculate the density of 'X' and the number of unit cell in 4g of 'X'. `(N_(A)=6.022xx10^(23)mol^(-1))`

Explanation

We know,
`rho=(ZxxM)/(N_(A)xxa^(3))`, where, `rho` = density of the unit cell, Z = no. of particle present in unit cell = 4 , M = atomic mass of the element = 40`g*mol^(-1)`, `N_(A)` = Avogadro's number = `6.022xx10^(23)`, a = edge length of the unit cell = 400pm = `400xx10^(-10)cm`.
`thereforerho=(ZxxM)/(N_(A)xxa^(3))=(4xx40)/(6.022xx10^(23)xx(400xx10^(_10))^(3))`
`=4.1514g*cm^(-3)`
Therefore, the number of unit cell in 4g of 'X'
`=(6.022xx10^(23)xx4)/(40xx4)=1.5055xx10^(22)`