Question

Silver crystallises in face-centred cubic lattice. If edge length of the unit cell is $4.07\times {10}^{-8}cm$ and density of silver is $10.48gc{m}^{-3}$, determine the relative atomic mass of silver.

Answer 1

d=Density =$\frac{Massofunitcell}{Volumeofunitcell}=\frac{2.M}{a^3{N}_A}$

$2=$ No. of atoms in one unit cell

$M=$ Mass of single atom

$a=$ Edge length

$N_A$= Avagadro Number

$M=\frac{d\times a^3\times Na}{2}$

For fcc lattice , 2=4

$M=\frac{\left(10.48\right)\times (4.07\times {10}^{-8}{)}^3\times (6.023\times {10}^{23}}{4}$

$=106.5$ grams.

Relative atomic mass of silver $=106.5$ grams.