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

`6.04xx10^23`

Explanation

For BCC unit cell of the element Fe, Z=2
`N_0=(ZxxM)/(a^3xxrho)=(2xx56 g mol^(-1))/((286.65xx10^(-10)cm)^3xx(7.87 g cm^(-3)))=6.04xx10^23 mol^(-1)`