Silver crystallises in F.C.C. (face-centred cubic crystal) structure. The edge length of the unit cell is found to be 408.7 pm. Calculate density of the unit cell.

[Given: Molar mass of silver is 108 g mol^{-1} ]

#### Solution

Given: Edge length (a) = 408.7 pm = 408.7 x 10^{–12} m = 408.7 x10^{-10} cm,

Molar mass/Atomic mass of silver = 108 g mol^{-1}

To find: Density (d)

Formulae:

`1."Mass of one atom"="Atomic mass"/"Avogardo Number"`

`2. Volume of unit cell=a^3`

`3. "Density"="mass of unit cell"/"Volume of unit cell" `

Calculation: For fcc lattice, number of atoms per unit cell is 4.

Mass of one atom of silver=`"Atomic mass"/"Avogadro number"`

`=108/(6.023xx10^23)=17.9xx10^(-23)g`

Mass of unit cell = 4 x 17.9 x 10^{–23} = 71.7 x 10^{–23} g

Volume of unit cell = a^{3} = (408.7 x 10^{–10} cm)^{3} = 6.827 x10^{–23} cm^{3}

`"Density"="mass of unit cell"/"Volume of unit cell" `

`=(71.7xx10^(-23)g)/(6.827xx10^(-23)cm^3)=10.5g `

#### Notes

The above numerical can also be solved using the formula, `d=(z.M)/(a^3N_A)`