Silver crystallises in fcc lattice.

Question:

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5 g cm−3, calculate the atomic mass of silver.

Solution:

It is given that the edge length, a = 4.077 × 10−8 cm

Density, d = 10.5 g cm−3

As the lattice is fcc type, the number of atoms per unit cell, z = 4

We also know that, NA = 6.022 × 1023 mol−1

Using the relation:

$d=\frac{z \mathrm{M}}{a^{3} \mathrm{~N}_{\mathrm{A}}}$

$\Rightarrow \mathrm{M}=\frac{d a^{3} \mathrm{~N}_{\mathrm{A}}}{z}$

$=\frac{10.5 \mathrm{gcm}^{-3} \times\left(4.077 \times 10^{-8} \mathrm{~cm}\right)^{3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}}{4}$

= 107.13 gmol−1

Therefore, atomic mass of silver = 107.13 u

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