An element crystallizes in a structure having fee unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains `24 xx 10^(23)` atoms.

Answer

41.67 g `cm^(-3)`

Explanation

a = 200 pm = `200 xx 10^(-10)` cm, Z= `4.24 xx 10^(23)` atoms of the element are present in 200 g
`6.022 xx 10^(23)` atoms of the element are present in mass `200 xx 6.022 xx 10^(23)` = 50.18 g
`:.` Atomic mass of the element = 50.18 g `mol^(-1)`
`d= (Z xx M)/(N_A xx a^3 "(in pm) "xx 10^(-30)) = (4 xx (50.18 g mol^(-)))/((6.022 xx 10^(23) mol^(-1)) xx (200)^3 x 10^(-30) cm^3) = 41.67 g cm^(-3)`