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. $5.0kg{m}^{-3}$
• B. $5.0g{m}^{-3}$
• C. $5.0kgc{m}^{-3}$
• D. $5.0gc{m}^{-3}$

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,
$\text{a}=2\left({\left(\text{r}\right)}_{{\left(\text{A}\right)}^{+}}+{\left(\text{r}\right)}_{{\left(\text{B}\right)}^{-}}\right)=2{\left(\text{Y}\right)}^{1/3}\times 10^{-9}\text{m}$
and N_o = Avogadro number = $6.023\times 10^{23}$
Density
$=\frac{4\times \left(6\cdot 023\text{Y}\times 10^{-3}\right)\text{Kg}{\left(\text{ mole}\right)}^{-1}}{6\cdot 023\times 10^{23}\times {\left(2{\left(\text{Y}\right)}^{1/3}\times 10^{-9}\right)}^3{\left(\text{m}\right)}^3{\left(\text{ mole}\right)}^{-1}}$
$=\frac{4\times 6\cdot 023\times 10^{-3}\times \text{Y}}{6\cdot 023\times 8\times \text{Y}\times 10^{-4}}$
$=5\text{ Kg }{\text{m}}^{-3}$