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An element crystallizes in a bcc lattice with a cell edge of \[3\times {{10}^{-8}}cm\] . The density of the element is $6.89gc{{m}^{-3}}$ . Calculate the molar mass of the element. ( ${{N}_{A}}=6.022\times {{10}^{23}}mo{{l}^{-1}}$ )


Answer
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Hint: We know that any crystal lattice is made up of very large unit cells and every lattice point is occupied by one constituent particle (atom, molecular, or ion). We must consider three types of cubic unit cells and for simplicity assume that the constituent particle is an atom, which is a primitive cubic unit cell, body-centered cubic unit cell, and face-centered cubic unit cell.

Complete step by step solution:
 A Body-centred cubic (bcc) unit cell with an atom at each of its corners and also one atom at its body center. An atom at the body center can be seen that wholly belongs to the unit cell in which it is present.
Hence, in a bcc unit cell:
(i) $8~corners\times \dfrac{1}{8}per~~corner~atom=8\times \dfrac{1}{8}$ = 1 atom
(ii) 1 body centre atom = $1\times 1$ = 1 atom
Therefore, total number of atoms per unit cell for bcc, Z = 2 atoms
Given, edge of lattice (a) = $3\times {{10}^{-8}}cm$
Density of element (d) = $6.89gc{{m}^{-3}}$
${{N}_{A}}=6.022\times {{10}^{23}}mo{{l}^{-1}}$
Molar mass of element (M) = $\dfrac{d\times {{N}_{A}}\times {{a}^{3}}}{Z}$
M = $\dfrac{6.89\times 6.022\times {{10}^{23}}\times {{(3\times {{10}^{-8}})}^{3}}}{2}=56.013g\simeq 56g$

Hence, the molar mass of the element is 56g

Note: A cubic crystal is made up of a crystal lattice or a space lattice structure. A repeating arrangement of a space lattice is made up of unit cells. The most basic structure of a crystalline solid is a unit cell. The density of the unit cell is said to be the density of the cubic crystal itself.