An element A crystallises in fcc structure. 200 g of this element has `4.12xx10^(24)` atoms. If the density of A is `7.2"g cm"^(-3)`, calculate the edge length of the unit cell.

Explanation

We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))or a^(3)=(ZxxM)/(rhoxxN_(0)xx10^(-30))`
According to available data,
No. of atoms in the fcc unit cell (Z) = 4
Density of unit cell `(rho)=7.2"g cm"^(-3)`
Atomic mass (M) of the element may be calculated as follows :
`4.12xx10^(24)` atoms of the element are present in 200 g
`6.022xx10^(23)` atoms of the element are present in `=((200g))/(4.12xx10^(24))xx6.022xx10^(23)=29.23g`
Atomic mass (M) of the element = `29.23"g mol"^(-1)`
Avogadro's number `(N_(0))=6.022xx10^(23)mol^(-1)`
By substituting the values, `a^(3)=(4xx("29.23 g mol"^(-1)))/((7.2"g cm")^(-3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)"cm"^(3)))`
`=26.97xx10^(6)("pm")^(3)" "(because"value of a is in pm")`
Edge Length (a) =`[26.97xx10^(6)("pm")^(3)]^(1//3)=2.999xx10^(2)"pm"=299.9"pm"`