Question

A crystal of lead (II) sulphide has NaCl structure. In this crystal the shortest distance between a $P{b}^{2+}$ ion and $S^{2-}$
ion is 300 pm. What is the volume of a unit cell of lead sulphide ?

• A. $216\times {10}^{-24}c{m}^3$
• B. $2.7\times {10}^{-23}c{m}^3$
  
• C. $216\times {10}^{-30}c{m}^3$
  
• D. $27\times {10}^{-30}c{m}^3$

Answer 1

The correct option is A $216\times {10}^{-24}c{m}^3$

Side length of unit cell of crystal = $2\times 300=600pm$

${\text{vol = (side length)}}^3=(600{)}^{3}p{m}^3=216\times {10}^6\times {10}^{-30}c{m}^3=216\times {10}^{-24}c{m}^3$