An element (atomic mass $100 \ g/mol$) having $bcc$ structure has unit cell edge $400 \ pm$. Then density of the element is (in $g/cm^3$)

Solid $AB$ has $NaCl$ type structure. If the radius of $A^{+}$ and $B^{-}$ are $0.5 \ \mathring{A}$ and $1.5 \ \mathring{A}$ respectively and the formula mass of $AB$ is $48 \ g/mol$,what is the density of $AB$ solid? $(N_A = 6 \times 10^{23})$

Niobium crystallises in a body-centred cubic $(bcc)$ structure. If the density is $8.55 \ g \ cm^{-3}$, then the atomic radius of niobium is (atomic mass of niobium $= 93 \ u$) (in $pm$)

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 $.9$)

$Fe$ crystallizes in a $bcc$ lattice. If the edge length is $286.65 \,pm$ and the density is $7.874 \,g/cm^3$, calculate the Avogadro number. $(Fe = 55.845 \,u)$

Calculate the number of unit cells in $38.6 \ g$ of noble metal having density $19.3 \ g \ cm^{-3}$ and volume of one unit cell is $6.18 \times 10^{-23} \ cm^{3}$?