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)

Face-centred cubic lattice

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.