An element of atomic number 52 occurs in bcc structure with cell edge length of 288 pm. Calculate the Avodadro's number if density is `7.2gm//cm^(3)`.

Answer

`6.05xx10^(23)mol^(-1)`

Explanation

`"Density"(rho)=(zxxM)/(a^(3)xxN_(0)xx10^(-30))orN_(0)=(zxxM)/(a^(3)xxrhoxx10^(-30))`
Edge length (a) = 288 pm = 288 , Atomic mass of element (M) = 52 g `mol^(-1)`
No. of atoms per unit cell (z) = 2 , Density of the element `(rho)=7.2"g cm"^(-3)`
Avogadro's number `(N_(0))=(2xx("52 g mol"^(-1)))/((288)^(3)xx(7.2"g cm"^(-3))xx(10^(-30)cm^(3)))=6.05xx10^(23)"mol"^(-1)`