The density of KBr is 2. 75 gm/cc length of the unit cell is 654 pm, K = 38, Br = 80, then what is true about the predicted nature of the solid?

The expression for the density of the unit cell is as given below: 

$\mathrm{d}=\frac{\mathrm{Z}\times \mathrm{M}}{\mathrm{V}\times {\mathrm{N}}_{\mathrm{o}}}=\frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{o}}}$

Hence, $\mathrm{Z}=\frac{\mathrm{d}\times {\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{o}}}{\mathrm{M}}$

Substituting values in the above expression, we get 

$\mathrm{z}=\frac{2.75\times {\left(654\times {10}^{-10}\right)}^3\times 6.023\times {10}^{23}}{119}\Rightarrow \mathrm{n}=4$

The unit cell contains 4 molecules of KBr.

Hence, it is a face-centered cubic unit cell.

Hence, the correct option is A.