Question

An element crystallizes in fcc lattice with a cell of edge length 300 pm. The density of the element is 10.8 g cm${}^{–3}$. Calculate the number of atoms in 108 g of the element.

Answer 1

Given that,

An element crystallizes in a fcc lattice with a cell edge length of 300 pm.

Density of the element $=$ 10.8 g.cm${}^{-3}$

To find,

Number of atoms in 108 g of the element

We know,

Density, $d=\frac{z\times M}{N_A\times a^3}$

"or"
$N=\frac{z\times M}{d\times a^3}$ : Note: Here, $z=4$ since it is fcc unit cell.

$N=\frac{4\times 108}{10.8\times (3\times {10}^{-8}{)}^3}=1.48\times {10}^{24}atoms$