Question

The density of KBr is $2.75 \, g \, cm^{- 3}$ , 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 $\rho=2.75 g cm ^{-3}$
we need to find type of unit cell
$\therefore \rho=\frac{Z \times M}{N_A a^3} $
$\because \quad 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 .