Question

An ionic compound $\text{AB}$ has a rock salt structure with $\text{A:B = 1:1}$. The formula mass of $\text{AB}$ is $6.023\text{y}$ amu and the closest $\text{A - B}$ distance is $y^{\frac{1}{3}}\text{nm}$. Calculate the density of the lattice in ${\text{kg/m}}^3$.

Answer 1

Given distance between A and B = $y^{1/3}nm=y^{1/3}\times {10}^{-9}m$
Since we have been told that it is a rock salt type structure, we understand that the above distance taken twice will be the edge length.
Hence, edge length = $2y^{1/3}\times {10}^{-9}m$
We also know that $Z=4$
We know that density = $\rho =\frac{ZM}{V{N}_A}=\frac{4\times 6.023y\times {10}^{-3}}{(2\times y^{1/3}\times {10}^{-9}{)}^3\times 6.023\times {10}^{23}}=5kg/m^3$