Question

Lead sulphide has a face-centered 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. of Pb = 207, S = 32)

Answer 1

`rho = ("Z" xx "M")/("N"_0"a"^3)`

a = 495 × 10^-10 cm

M = 207 + 32 = 239 U

N₀ = 6.022 × 10²³

Z = 4

`rho = (4 xx 239)/(6.022 xx 10^23 xx (4.95)^3 xx 10^-24)`

`= (4 xx 2390)/(6.022 xx (4.95)^3)`

`= (4 xx 2390)/(6.022 xx 121.29)` 

`= 7560/730.41`