Calculate the radius of an atom of metal if it forms a simple cubic unit cell with an edge length of $380 \ pm$. (in $pm$)

Calculate the volume of the unit cell for an element having a molar mass of $92 \ g \ mol^{-1}$ that forms a $bcc$ structure,given $\left[\varrho \times N_{A} = 5.0 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}\right]$.

The cubic unit cell of a metal (molar mass $= 63.55 \ g \ mol^{-1}$) has an edge length of $362 \ pm$. Its density is $8.92 \ g \ cm^{-3}$. The type of unit cell is

The edge length of the unit cell of a metal $(M_W = 24 \, g \, mol^{-1})$ having a cubic structure is $4.53 \, \mathring{A}$. If the density of the metal is $1.74 \, g \, cm^{-3}$,then the effective number of atoms in the unit cell is :- $(N_A = 6 \times 10^{23} \, mol^{-1})$

Iron exhibits $bcc$ structure at room temperature. Above $500^{\circ} C$,it transforms to $fcc$ structure. Find the ratio of the density of iron at room temperature to that at $500^{\circ} C$. (Assume the atomic radii and the molar mass of iron remain constant even with variation in temperature)

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length $405 \ pm$. If its density is $2.7 \times 10^{3} \ kg \ m^{-3},$ the radius of the element is approximately......... $\times 10^{-12} \ m$ (to the nearest integer).