Question

Lead sulphide has a 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. 
(Atomic weight of Pb = 207, S = 32)

Answer 1

Given: a = 495 × 10⁻¹⁰ cm

= 4.95 × 10⁻⁸ cm

M = 207 + 32 = 239 U

N_A = 6.022 × 10²³

Z = 4

Formula: `rho = (Z xx M)/(N_A xx a^3)`

`rho = (4 xx 239)/((4.95 xx 10^-8)^3 xx 6.022 xx 10^23)`

`rho = 956/(121.2 xx 10^-24 xx 6.022 xx 10^23)`

= `956/(1.21 xx 10^-22 xx 6.022 xx 10^23)`

= `956/(7.29 xx 10^1)` 

= `956/72.9` 

= 13.1 g/cm³