Question

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

Answer 1

Given, $d=10.5gc{m}^{-3}$

$a=4.07\times {10}^{-8}cm$

Z = 4(fcc)atoms

$N_A=6.022\times {10}^{23}mo{l}^{-1}$ (Avogadro's constant)

$M=\frac{d\times a^3\times N_A}{Z}$

$\frac{10.5gc{m}^{-3}\times (4.077\times {10}^{-8}cm{)}^3\times (6.022\times {10}^{23}mo{l}^{-1})}{4}$

$=\frac{10.5\times 67.767\times {10}^{-24}\times 6.022\times {10}^{23}}{4}gmo{l}^{-1}$

$=107.09gmo{l}^{-1}$