Silver crystallises in fcc lattice. If edge length of the cell is `4.07 xx 10^(-8)` cm and density is `10.5 "g cm"^(-3)`, calculate the atomic mass of silver. 

Explanation

Edge length `(a) = 4.07 xx 10^(-8)` cm
`therefore` Volume of unit cell `= a^3 = (4.07)^3 xx 10^(-24) "cm"^3`
Density `(d) = 10.5 "g cm"^3`
Atomic Mass (M) = ?
Number of atoms per unit cell (Z) = 4
`therefore M= (d xx N_A xx a^3)/(Z)`
`therefore M = (10.5 xx 6.022 xx 10^23 xx (4.07)^3 xx 10(-24))/(4)`
`therefore` M = 107.12 g/mol .