An element crystallizes in fcc lattice. If the edge length of the unit cell is 408.6 pm and the density is `10.5 g cm^(-3)`. Calculate the atomic mass of the element.

Explanation

`a=408.6 xx 10^(-12)m = 408.6 xx 10^(-10)cm`.
`d=10.5 g cm^(-3)`
`N_(A)= 6.022 xx 10^(23)"mol"^(-1)`
Z = 4
`:.` The crystal is fcc lattice
`M=(da^(3)N_(A))/(Z)=(10.5 g cm^(-3)xx (408.6xx10^(-10) cm)^(3) xx 6.022 xx 10^(23) "mol"^(-1))/(4)`
`=107.8"g mol"^(-1)`
* Atomic mass of the element = 107.8 u.