The nearest neighbour silver atoms in silver crystal are `2.87xx10^(-10)` m apart. What is the density of the silver metal ? Silver crystallises in fcc form. (Atomic mass of Ag = 108)

Answer

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

Explanation

Density `(rho)=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))`
For fcc structure edge length (a) = `sqrt2xx"distance"=1.414xx2.87xx10^(-10)m=" 406 pm 406"`
Atomic mass of Ag (M) = 108 g `mol^(-1)`
No. of atoms per unit cell (z) = 4 , Avogadro's no. `(N_(0))=6.022xx10^(23)mol^(-1)`
`therefore" Density"(rho)=(4xx("108 g mol"^(-1)))/((406)^(3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)xxcm^(3)))=10.72"g cm"^(-3)`