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 $.9$)

Calculate the density of an element with molar mass $27 \ g \ mol^{-1}$ having $4$ atoms in a unit cell with edge length $405 \ pm$. (in $g \ cm^{-3}$)

The density of $NaCl$ is $2.165 \, g \, cm^{-3}$. The distance between $Na^+$ and $Cl^-$ ions in the $NaCl$ crystal is $281 \, pm$. Based on this information,the value of Avogadro's number is ...

An element $M$ crystallises in a body-centred cubic unit cell with a cell edge of $300 \, pm$. The density of the element is $6.0 \, g \, cm^{-3}$. The number of atoms present in $180 \, g$ of the element is $............ \times 10^{23}$ (Nearest integer).

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

Calculate the number of unit cells in $0.79 \ g$ of a metal if the product of the density and the volume of the unit cell is $1.58 \times 10^{-22} \ g$.