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

`z=4, d = 5.2 g//cm^(3), M = ?`
`a = 300 p m = 3xx10^(-10)m=3xx10^(-8)cm`
`a^(3)=27xx10^(-24)cm^(3)`
`N_(o)=6.023xx10^(23)`
`M=(rho N_(o)a^(3))/(z)=(5.2xx6.023xx10^(23)xx27xx10^(-24))/(4)`
= 21.13 g/mol
`6.023xx10^(23)` atoms have = 21.13 g
`therefore 3.01xx10^(24)` atoms have `= (21.13xx3.01xx10^(24))/(6.02xx10^(23))`
`= 105.65 g`