Question

The face-centered unit cell of nickel has an edge length of 352.39 pm. The density of nickel is 8.9 $gc{{m}^{-3}}$. Calculate the value of Avogadro's number. The atomic mass of nickel is 58.7 and 1 pm is equal to ${{10}^{-10}}$ cm.

Answer 1

Hint: In this question we have to find the value of the avogadro's number with the help of edge length and density. To solve this question we should have prior knowledge about FCC and the relation between the edge length and density in FCC.

Complete step by step solution:
Face centered cubic cell like all the lattices has a lattice point at the eight corner of the unit cell plus added to the additional points at the center of each face of the unit cell. The simplest crystal structure are the crystals in which there is only a single atom present at each lattice point in the system.
The number of atoms present in FCC is 4.
Given in the question:
The face-centered unit cell of nickel has an edge length = 352.39 pm
The face-centered unit cell of nickel has a density = 8.9 $gc{{m}^{-3}}$
The atomic mass of nickel is = 58.7 and 1 pm is equal to ${{10}^{-10}}$ cm
Density = $\delta =\dfrac{(Z)(M)}{{{a}^{3}}{{N}_{A}}}$
$\begin{align}
& {{N}_{A}}=\dfrac{(Z)(M)}{\delta {{N}_{A}}}=\dfrac{(4)(58.7)}{(8.9){{(352.39({{10}^{-10}}))}^{3}}} \\
 & \Rightarrow {{N}_{A}}=(6.029)({{10}^{23}}) \\
\end{align}$

Hence the avogadro's number = $(6.029)({{10}^{23}})$

Note: In the face centered unit cell, atoms are present at the face and at the corner of the cube. And as we know that there are 6 faces present at 6 corners and each atom contributes half to the unit cell in a cube hence 3 atoms present and one atom at the corner contributes to eight other corners. So the total number of atoms present in the body centered cubic cell is = 4. i.e.
3 atom at faces $+\text{ (}\dfrac{1}{8})(8)$ at the corner = 4 atoms