Calculate the number of atoms in $0.3 \ g$ of a metal if it forms a $bcc$ structure,given that $[\rho \times a^3 = 3 \times 10^{-22} \ g]$.

Niobium crystallizes in a body-centered cubic $(BCC)$ structure. If the density is $8.55 \ g \ cm^{-3}$,calculate the atomic radius of niobium. (Atomic mass of niobium $M_w = 93 \ g \ mol^{-1}$)

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).

$A$ compound can crystallise in two forms $\alpha$ and $\beta$ which are $fcc$ and $bcc$, respectively. The $\alpha$-form has a side length of $2 \ pm$ and the $\beta$-form has a side length of $4 \ pm$. The ratio of their density $\frac{\rho_\alpha}{\rho_\beta}$ is

Calculate the number of atoms present in $1 \text{ g}$ of a metal that forms a simple cubic unit cell structure, given that the product of density and volume of the unit cell is $66.5 \times 10^{-24} \text{ g}$.

An element with $BCC$ structure has an edge length of $500 \ pm$. If its density is $4 \ g \ cm^{-3}$, find the atomic mass of the element.