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 $pm$.

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length of $405 \ pm$. If its density is $2.7 \times 10^3 \ kg \ m^{-3}$, the nature of the cubic unit cell is: $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$

The correct relationship between unit cell edge length '$a$' and radius of sphere '$r$' for face-centred and body-centred cubic structures respectively are:

$CsBr$ crystallises in a body-centred cubic lattice. The unit cell length is $436.6 \, pm$. Given that the atomic mass of $Cs = 133$ and that of $Br = 80 \, amu$ and Avogadro number being $6.02 \times 10^{23} \, mol^{-1}$,the density of $CsBr$ is .............. $g/cm^{3}$.

The density of a body-centered cubic $(BCC)$ crystal of Molybdenum is $10.3 \ g \ cm^{-3}$. Calculate the edge length of the unit cell in $pm$. (Atomic mass of $Mo = 95.94 \ g \ mol^{-1}$) (in $.9$)

Element $X$ crystallizes in a $12$ coordination face-centered cubic $(fcc)$ lattice. On applying high temperature,it changes to an $8$ coordination body-centered cubic $(bcc)$ lattice. Find the ratio of the density of the crystal lattice before and after applying high temperature. The atomic radius of $X$ is the same in both crystals.