Question

Silver crystallises in fcc lattice. If edge length of the cell is $4.07\times {10}^{-8}cm$ and density is $10.5g$ ${cm}^3$, calculate the atomic mass of silver.

Answer 1

Given, edge length (l)=$4.07\times {10}^{-8}$ cm

density (d)=10.5 g/cc

Number of atoms in fcc lattice=4

Avagadro's number=$6.023\times {10}^{23}$

Density$=d=\frac{ZM}{l^3{N}_A}$

$⟹M=\frac{d{l}^3{N}_A}{Z}=\frac{\left(10.5\right){\left(4.07\right)}^3(6.023\times {10}^{23})}{4}=107.09g{mol}^{-1}$