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=(ZxxA)/(rhoxxa^3)=(2xx52)/(7.2xx(288.4xx10^(-10))^3)`
`N_0=6.02xx10^23` atoms