The density of KBr is 2.75 g cm-1. The length of the edge of the unit cell is 654pm. What will be the type of cubic lattice to which unit cell of KBr belongs?
(Atomic mass of K= 39; Atomic mass of Br = 80)
Answer & explanation
Correct answer: option 2
The correct answer is option 2. Face-centred cubic lattice.
\(\text{For cubic crystals, }\rho\text{ = }\frac{Z×M}{a^3×N_o}\)
\(or, \text{ Z = }\frac{\rho \text{ × }a^3\text{ × }N_o}{M}\)
\(or, \text{ Z = }\frac{(2.75 \text{ g }cm^{-1})(654\text{ × }10^{-10}cm)^3(6.023\text{ × }10^{23}mol^{-1})}{(39\text{ + }80)g\text{ mol}^{-1}}\)
\(or, \text{ Z = 3.89 = 4}\)
Thus, there are four formula units of KBr present per unit cell. Hence, it has face-centred cubic lattice.