Question

A compound $AB$ has rock salt type structure. The formula weight of $AB$ is $6.023Y$ amu, and closest $A-B$ distance is $Y^{1/3}$ nm, where $Y$ is an arbitrary number, find the density of the lattice.

• A. $5$ kg$/m^3$
• B. $5\times {10}^{-3}$ g$/m^3$
• C. ${10}^{-3}$ g$/m^3$
• D. $15\times {10}^{-2}$ g$/m^3$

Answer 1

The correct option is A $5$ kg$/m^3$

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 $AB=\frac{n\times mol.wt}{N_{A}V}$

$=\frac{4\times 6.023\times y\times {10}^{-3}}{6.023\times {10}^{23}\times {\left(2\times \frac{y}{3}{10}^{-9}\right)}^3}$

$=5.0kg{m}^{-3}$.