Question

Lead sulphide has face centered cubic crystal structure. If the edge length of the unit cell of lead sulphide is $495pm$, calculate the density of the crystal. (atomic weight $Pb=207,S=32$)

Answer 1

Atomic weight, $Pb$=$207,S$=$32$

Molecular weight (M) of $PbS=207+32=239$ g/mol.

An fcc unit cell has 4 formula units (Z) per unit cell.
The edge length (a) of the unit cell of lead sulphide is 495pm or $495\times {10}^{-10}$ cm

The density $d=\frac{ZM}{a^3{N}_A}$

$d=\frac{4\times 239}{(495\times {10}^{-10}{)}^3\times 6.023\times {10}^{23}}$
$d=13.1g/c{m}^3$