The crystal of $CsBr$ has an edge length of $437 \ pm$. If the density of the crystal is $4.24 \ g \ cm^{-3}$, determine the type of crystal structure of $CsBr$ (Atomic mass of $Cs = 133, Br = 80$).

The face diagonal of a cubic close-packed unit cell is $4 \ \mathring{A}$. What will be the edge length?

$KBr$ has $NaCl$ structure. Its density is $2.75 \ g \ cm^{-3}$. The edge length of the unit cell will be (molar mass of $KBr$ is $119 \ g \ mol^{-1}$):

Calculate the molar mass of an element having density $7.8 \ g \ cm^{-3}$ that forms a $bcc$ unit cell. $\left[a^3 \cdot N_{A} = 16.2 \ cm^3 \ mol^{-1}\right]$

The density of $KBr$ is $2.75 \, g \, cm^{-3}$ and the unit cell edge length is $654 \, pm$. Given the atomic masses $K = 39$ and $Br = 80$, what is the crystal structure of $KBr$?

Ionic radii of cation $A^{+}$ and anion $B^{-}$ are $102 \ pm$ and $181 \ pm$ respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for $B^{-}$ and $A^{+}$ is present in all octahedral voids. The edge length of the unit cell of the crystal $AB$ is $pm$.