Question

Calculate the density of PbS crystal (fcc) if the edge length of its unit cell is 500pm. (NA = 6.022 x 1023

,

atomic mass of Pb = 207.2, S = 32)​

Answer 1

Given:

PbS forms FCC structure.

a = 500 pm = 5 × 10⁻⁸ cm

NA = 6.022 × 10²³

Atomic mass of Pb = 207.2 and S = 32

To Find:

The density (d) of PbS crystal.

Calculation:

- Molar mass of PbS = 207.2 + 32 = 239.2 gm

- For FCC, z = 4

- V = a³ = (5 × 10⁻⁸)³

⇒ V = 125 × 10⁻²⁴ cm³

- We know the formula:

V × NA × d = z × M

⇒ d = (z × M)/(V × NA)

⇒ d = (4 × 239.2)/(125 × 10⁻²⁴ × 6.022 × 10²³)

⇒ d = 956.8 / (752.75 × 10⁻¹)

⇒ d = 12.71 gm/cm³

- So, the density of PbS is 12.71 gm/cm³.