The compound CuCl has ZnS structure and the edge length of the unit cell is 500 pm. Calculate its density (Atomic mass of Cu = 63, Cl = 35.5)

Answer

`5.23"g cm"^(-3)`

Explanation

Density `(rho)=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))`
Edge length (a) = 500 pm 500
Gram formula mass of CuCl (M) = `63+35.5=98.5 "g mol"^(-1)`
No. of formula units of unit cell (z) = 4 , Avogadro's No. `(N_(0))=6.022xx10^(23)"mol"^(-1)`
`"Density"(rho)=(4xx("98.5 g mol"^(-1)))/((500)^(3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)xx"cm"^(3)))=5.23"g cm"^(-3)`