Silver crystallines in fcc lattice. If edge length of the cell is `4.07 xx 10^(-8) cm` and density is 10.5 `gcm^(-3)` . Calculate the atomic mass of silver.

Explanation

`M =( d xx a^(3) xx NA)/( g)`
d= Density of the material
a= Length of the edge of the cell
`N_(A) = ` Avagadro number
Z= No. of atoms
`M = (( 6.023 xx 10^(23) mol^(-1)))/(4)`
Atomic mass of silver `M = 107.08 g mol^(-1)`