$CsCl$ has a $bcc$ arrangement. Its unit cell edge length is $400 \, pm$. What is its inter-ionic distance?

$A$ given metal crystallises out with a cubic structure having edge length of $361 \, pm$. If there are four metal atoms in one unit cell,what is the radius of one atom? .............. $pm$

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

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 ...

Calculate the number of unit cells in $58.5 \, g$ of $NaCl$ crystallizing in an $fcc$ structure. $(NaCl = 58.5 \, g/mol)$

How many atoms of niobium are present in $2.43 \ g$ if it forms $bcc$ structure with density $9 \ g \ cm^{-3}$ and volume of unit cell $2.7 \times 10^{-23} \ cm^3$?