Sodium metal crystallizes in a $bcc$ lattice with a unit cell edge length of $a = 4.29 \ \mathring{A}$. What is the radius of the sodium atom in $\mathring{A}$?

An element crystallises in $bcc$ type having atomic radius $1.33 \times 10^{-8} \ cm$,the edge length of the unit cell will be:

Calculate the density of an element having molar mass $27 \ g \ mol^{-1}$ that forms $fcc$ unit cell. $[a^3 \cdot N_A = 38.5 \ cm^3 \ mol^{-1}]$ (in $g \ cm^{-3}$)

Find the radius of a metal atom in a $bcc$ unit cell having an edge length of $450 \ pm$. (in $pm$)

$A$ metal has a $fcc$ lattice. The edge length of the unit cell is $404 \, pm$. The density of the metal is $2.72 \, g \, cm^{-3}$. The molar mass of the metal is ................. $g \, mol^{-1}$.
($N_A$ Avogadro's constant $= 6.02 \times 10^{23} \, mol^{-1}$)

$A$ face-centered cubic $(FCC)$ solid of an element (atomic mass $60$) has a cubic edge length of $4 \times 10^{-8} \, cm$. If Avogadro's number is $6 \times 10^{23} \, mol^{-1}$,then the density of the solid is: