Question

A compound AB has a rock salt type structure with A : B = 1 : 1. The formula weight of AB is 6.023 y g and the closest A - B distance is $y^{1/3}$ nm. If the observed density of lattice is found to be $20kg{m}^{-3}$, the density of lattice is

• A. $1.3kg{m}^{-3}$
• B. $2.13kg{m}^{-3}$
• C. $3.01kg{m}^{-3}$
• D. $5.0kg{m}^{-3}$

Answer 1

Answer: (D)

Let $AB$ has rock salt structure $(A:B::1:1)$
For F.C.C structure $(n=4)$ and formula weight of $AB$ is $6.023$ y g
Closest distance $A-B=y\frac{1}{3}nm$
Therefore,
Edge length of unit cell,
$a=2\left(A^{+}{B}^{-}\right)=2\times \frac{y}{3}\times 10^{-9}m$
Therefore,
Density of $AB=\frac{n\times mol.wt}{N_{A}V}$
$\begin{array}{l}=\frac{4\times 6.023\times y\times 10^{-3}}{6.023\times 10^{23}\times {\left(2\times \frac{2}{3}10^{-9}\right)}^3}\\ =5.0kg{m}^{-3}\end{array}$