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 $20\, kg\, m^{-3}$, the density of lattice is

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

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 $A B$ 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}} \\ =5.0 kg m ^{-3} \end{array} $