Iron has body centred cubic cell with a cell edge of 286.5 pm. The density of iron is 7.87 g `cm^(-3)`. Use this information to calculate Avogadro's number. (Atomic mass of Fe = 56 `mol^(-3)`)

Answer

`6.043xx10^(23)mol^(-1)`

Explanation

We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30)) or N_(0)=(ZxxM)/(a^(3)xxrho xx10^(-30))`
Edge length of the unit cell (a) = 286.5 pm = 286.5
No. of atoms per unit cell (Z) = 2
Density of unit cell `(rho)=7.87gcm^(-3)`
Atomic mass of iron (M) = 56 `mol^(-1)`
Avogadro's Number `N_(0)=(2xx(56gmol^(-1)))/((286.5)^(3)xx(7.87gcm^(-3))xx(10^(-30)cm^(3)))=6.043xx10^(23)mol^(-1)`