The element chromium crystallises in a body centred cubic lattice whose density is `7.20g//cm^(3)`. The length of the edge of the unit cell is 288.4 pm. Calculate Avogadro's number (Atomic mas of Cr = 52).

Explanation

For bcc, Z = 2, `rho = 7.2 g//cm^(3)`
a = 288.4 pm or `(288.4xx10^(-10))cm`
A = 52
`N_(0)=?`
`N_(0)=(Z xx A)/(rho a^(3))=(2xx52)/(7.2xx(288.4xx10^(-10))^(3))`
`N_(0)=6.02xx10^(23)` atoms