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).

Answer

`6.02 xx 10^(23)`

Explanation

If `N_A` is Avogadro number
Edge of unit cell =288.4 pm = `288.4 xx 10^(-10) cm`
`Z = 2` for bcc
Density = `(Z xx M)/(a^3 xx N_A)`
`:. 7.20 = (2 xx 52)/((288.4 xx 10^(-10))^(3) xx N_A)`
or `N_A = (2 xx 52)/(7.20 xx (288.4 xx 10^(-10))^(3))`
`= 6.02 xx 10^(23)`.