Question

A compound $AB$ having formula weight $=6.023\times YAMU$ , has rock salt type structure and the closest AB distance is $Y^{\frac{1}{3}}$ nm, where $Y$ is an arbitrary number. The density of lattice is

• A. $5.0 \, kg \, m^{- 3}$
• B. $5.0 \, g \, m^{- 3}$
• C. $5.0kg \, cm^{- 3}$
• D. $5.0g \, cm^{- 3}$

Answer 1

Answer: (A)

$\text{Density of AB = } \frac{Z \times M}{N_{0} \times a^{3}}$
$Here, \, Z=4$ (for fcc ), M=6.023 Y,
$\text{a} = 2 \left(\left(\text{r}\right)_{\left(\text{A}\right)^{+}} + \left(\text{r}\right)_{\left(\text{B}\right)^{-}}\right) = 2 \left(\text{Y}\right)^{1 / 3} \times 1 0^{- 9} \text{m}$
and N_o = Avogadro number = $6.023 \, \times 10^{23} \, $
Density
$= \frac{4 \times \left(6 \cdot 0 2 3 \text{Y} \times 1 0^{- 3}\right) \text{Kg} \left(\text{ mole}\right)^{- 1}}{6 \cdot 0 2 3 \times 1 0^{2 3} \times \left(2 \left(\text{Y}\right)^{1 / 3} \times 1 0^{- 9}\right)^{3} \left(\text{m}\right)^{3} \left(\text{ mole}\right)^{- 1}}$
$= \frac{4 \times 6 \cdot 0 2 3 \times 1 0^{- 3} \times \text{Y}}{6 \cdot 0 2 3 \times 8 \times \text{Y} \times 1 0^{- 4}}$
$= 5 \text{ Kg } \text{m}^{- 3}$