Question

The density of KBr is $2.75gc{m}^{-3}$ , length of the unit cell is 654 pm. K = 39, Br = 80, then what can be true about the predicted nature of the solid?

(Given: ${\text{N}}_{\text{A}}=6.023\times 10^{23}$ )

• A. Solid has face centred cubic system with co-ordination number = 6
• B. Solid has simple cubic system with co-ordination number = 8
• C. Solid has face centred cubic system with co-ordination number =12
• D. None of the above

Answer 1

Answer: (A)

Given that $\rho =2.75gc{m}^{-3}$
we need to find type of unit cell
$\therefore \rho =\frac{Z\times M}{N_A{a}^3}$
$\because Z=\frac{\rho \times N_A\times a^3}{M}$
$=\frac{2.75\times 6.023\times 10^{23}\times {\left(654\times 10^{-10}\right)}^3}{119}$
$\therefore z\simeq 4$ So fcc unit cell
In FCC 1 atm surrounded by 6 atoms so coordination number is 6 .