Silver crystallises in an fcc lattice. The edge length of its unit cell is `4.077 xx 10^(-8)` cm and its density is `10.5 g cm^(-3)`. Calculate on this basis the atomic mass of silver. `[N_(A)= 6.02 xx 10^(23) mol^(-1)]`

Explanation

Apply the relation :
`=(dxxa^(3)xxN_(A))/(z)`
`"Given : d = 10.5 g cm"^(-3), a=4.077xx10^(-8)cm`
In fcc lattice, z = 4
Substituting the values in the above equation, we get
`M=(10.5xx(4.077)^(3)xx10^(-24)xx6.02xx10^(23))/(4)=107.09`
Thus, the atomic mass of silver = 107.09u.