Question

A compound $AB$ has a rock salt type structure with $A:B=1:1$. The formula weight of $AB$ is $6.023y\text{g}$ and the closest $A-B$ distance is $y^{1/3}\text{nm}$. The density of lattice $\left({\text{in kg m}}^{-3}\right)$ is

Answer 1

$AB$ has rock salt $(A:B=1:1)$ structure, i.e., f.c.c. structure $(n=4)$ and formula weight of $AB$ is $6.023y\text{amu}$ having closest distance $A-B{y}^{1/3}\text{nm}$. Therefore edge lenghth of unit cell $2(A^{+}+B^{-})=2\times y^{1/3}\times {10}^{-9}\text{m}$
$\therefore$ Density of $AB=\frac{n\times \text{mol wt.}}{N_A\times V}=\frac{4\times 6.023\times y\times {10}^{-3}}{6.023\times {10}^{23}\times (2y^{1/3}\times {10}^{-9}{)}^3}=5.0{\text{kg m}}^{-3}$