Question

A compound $AB$ has a rock type structure with $A:B=1:1$. The formula mass of $AB$ is $6.023$ $y^{1/3}$ y amu and the closest $A-B$ distance is $y^{1/3}$ nm.Calculate the density of lattice.

• A. $5.0kg/m^3$
• B. $10kg/m^3$
• C. $50kg/m^3$
• D. $20kg/m^3$

Answer 1

Answer: (A)

$5.0kg/m^3$

$AB$ has rock-salt $(A:B::1:1)$ structure, i.e., fcc structure $(n=4)$ and formula mass of $AB$ is 6.023 y g having closest distance $A-B{y}^{1/3}$ nm. Therefore, edge length of unit cell
$\left(a\right)=2(A^{+}+B^{-})=2\times y^{1/3}\times {10}^{-9}m$
$\therefore$ $DensityofAB=\frac{n\times formulamass}{Av.no.\times V}=\frac{n\times formulamass}{6.023\times {10}^{23}\times a^3}$
$=\frac{4\times 6.023\times y\times {10}^{-3}}{6.023\times {10}^{23}\times (2y1/3\times {10}^{-9}{)}^3}$ (formula mass in kg)
$=5.0kg/m^3$