Silver crystallises in fcc lattice. If edge length of the cell is `4.07xx10^(-8)` cm and density is `10.5g*cm^(-3)`, calculate the atomic mass of silver.

Explanation

We know, `rho=(ZxxM)/(Nxxa^(3))`
Given: `a=4.07xx10^(-8)cm,rho=10.5g*cm^(-3)`
For fcc lattice, Z = 4
`therefore" "M=(rhoxxNxxa^(3))/Z=(10.5xx6.022xx10^(23)xx(4.07xx10^(-8))^(3))/4`
`=106.6g*mol^(-1)`
`therefore" "` Atomic mass of silver = `106.6g*mol^(1)`