Question

A compound $AB$ has (rock salt type structure) molecular weight of $AB$ is $6.023Yamu$ and closest $A-B$ distance is $Y^{1/3}$ nm, where $Y$ is an arbitrary number (in amu). Find the density of lattice, if the density of lattice is found to be $20kg{m}^{-3}$.

Answer 1

Fcc structure (n=4) and formula weight of AB is 6.023y g

closest distance $A-B=y\frac{1}{3}nm$

Therefore , edge length of unit cell,a $=2(A^{+}+B^{-})=2\times y/3\times {10}^{-9}m$

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

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

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