Question

An element crystallizes in a f.c.c lattice with a cell edge of $250$ pm. Calculate the density, if $300$ g of this element contains $2 \times {10^{24}}$ atoms.

Answer 1

Hint: The crystalline solids acquired a three dimensional arrangement of constituent particles which may be atoms, ions or molecules. If we represent each particle of the solid as a point diagrammatically, then such a type of arrangement is known as crystal lattice.
Each lattice is made up of unit cells. Unit cell is defined as the smallest portion of any lattice which replicates itself in three dimensions to make a complete crystal lattice.
Formula used :
To calculate the density of element is \[D = \dfrac{{Z \times M}}{{{a^3} \times {N_0}}}\],
Where, $D$ is density of unit cell
$Z$ is number of atom per unit of unit cell
$M$ is atomic mass of the given element
$a$ is length of the edge of the unit cell
${N_0}$ is Avogadro number with constant value

Complete answer:
Element crystallizes in the face center cell which means that there is one atom at each corner of the unit cell. Since, there are $8$ corners in any unit cell there must be 8 atoms of element presented.
Each atom at the corner contributes only ${\dfrac{1}{8}^{th}}$ to each unit cell because it is shared by $8$ cells.
Hence, the number of atoms of given element present at corners per unit cell is $ = 8$ corner atoms $ \times {\dfrac{1}{8}^{th}}$ atom per unit cell of f.c.c$ = 1$.
As we know that each unit cell has $6$ faces and each atom is present at the face of the unit cell by making $\dfrac{1}{2}$ contribution to each unit cell.
Hence, the number of atom of given element present at face per unit cell is $ = 6$ atoms at each faces $ \times \dfrac{1}{2}$ atom per unit cell of f.c.c $ = 3$
From the above calculation we find that the total number of atoms present at each f.c.c unit cell is $1 + 3 = 4$.
Hence, the value of $Z$ for f.c.c is $4$.
Now calculate the Molecular mass of the given element if $300$ g of element contains $2 \times {10^{24}}$ atoms.
To determine the Molecular mass of any element, multiply the Avogadro number with atomic weight of the element and number of atoms present in the particular weight of element.
$M = \dfrac{{300 \times 6.023 \times {{10}^{23}}}}{{2 \times {{10}^{23}}}}$
$M = 90.34$ g
Edge length of the unit cell is given $250$ pm.
Since, we know that $1$ cm $ = {10^{ - 8}}$ pm $ = {10^{ - 8}}$
So to convert the edge length of a unit cell in cm from pm multiply it by ${10^{ - 8}}$.
Now we get edge length $a = 250 \times {10^{ - 8}}$ cm.
Now substitute all the values in the formula of density, we get,
$D = \dfrac{{Z \times M}}{{{a^3} \times {N_0}}}$
$D = \dfrac{{4 \times 90.34}}{{{{\left( {2.5} \right)}^3} \times {{10}^{24}} \times 6.023 \times {{10}^{23}}}}$
$D = 38.4$ g/cm3

Note:
f.c.c comes under the category of non-primitive unit cells which are defined as “Unit cell which contain particles at corners of lattice as well as any other positions than of lattice".
Density of a unit cell is always the same as the density of a substance.
Convert all the units in the CGS system which means that the edge length of the unit cell must be in Centimeters, mass of element in grams.
$Z$ value of a simple cubic cell is $1$ while the body center cell is $2$.