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

We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))orN_(0)=(ZxxM)/(a^(3)xxrhoxx10^(-30))`
Edge length of the unit cell (a) = 288.4 pm = 288.4
No. of atoms per unit cell (Z) = 2
Density of the unit cell `(rho)=7.2g//cm^(3)`
Atomic mass of the element (M) = `52"g mol"^(-1)`
Avogadro's Number `(N_(0))=(2xx(52"g mol"^(-1)))/((288.4)^(3)xx(7.2"g mol"^(-3))xx(10^(-30)cm^(3)))=6.04xx10^(23)"mol"^(-1)`