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

Explanation

Edge length of unit cell
(a) = 495 pm
`=495 xx 10^(-10)` cm
Formula mass of lead sulphide (PbS),
M = 239.3 g/mol
No. of atoms in fcc unit cell
(Z) = 4
`N_(A)`, Avogadro number `=6.023xx10^(23)`
Density of crystal
`(d)=(Z xx M)/(a^(3)xx N_(A))`
`=(4xx239.3)/((495xx10^(-10))^(3))xx 6.023 xx 10^(23)`
`d=13.1g//cm^(3)`