The crystal structure of an element has an $fcc$ lattice. If the edge length of the crystal is $4 \ \mathring{A}$,what is the atomic weight (in $g \ mol^{-1}$) of the element,if the density of the crystal is $11.21 \ g \ cm^{-3}$ $(N_{A} = 6.023 \times 10^{23} \ mol^{-1})$?

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 $.9$)

The volume occupied by a single $CsCl$ ion pair in a crystal is $7.014 \times 10^{-23} \ cm^{3}$. The smallest $Cs^{+}-Cs^{+}$ inter-nuclear distance is equal to the length of the side of the cube corresponding to the volume of one $CsCl$ ion pair. The smallest $Cs^{+}-Cs^{+}$ inter-nuclear distance is nearly:

If the density of a $CsCl$ crystal,which crystallizes in a $bcc$ structure,is $3.988 \, g/cm^3$,then the edge length of the unit cell will be ........ $(CsCl = 168.4 \, g/mol)$.

Silver crystallises in $fcc$ lattice. If edge length of the cell is $4.07 \times 10^{-8} \ cm$ and density is $10.5 \ g \ cm^{-3}$,calculate the atomic mass of silver.

In an $fcc$ lattice,the edge length of a silver unit cell is $4.077 \times 10^{-8} \ cm$ and the density is $10.5 \ g \ cm^{-3}$. Calculate the atomic mass of silver.