Question

An element crystallizes in fcc structure with a unit cell edge length of 200pm. Calculate its density if 200g of this element contains $24 \times {10^{23}}atoms$?

Answer 1

Hint: The full form of “FCC” is a face centered cubic unit cell where the total number of atoms per unit cell will be four atoms. To solve this question, we should know the formula to calculate the density of the unit cell.

Complete step by step answer:
The density of unit cell is given by:
\[d = \dfrac{{zM}}{{{a^3}{N_A}}}\]……………………Equation 1
Where, d = density of the unit cell
Z = The number of atoms present in one unit cell
${N_A}$= Avogadro constant
a = length of the unit cell
M = molar mass
The mass of the unit cell is given by :
\[m = \dfrac{M}{{{N_A}}}\]…………………….Equation 2
\[M = m \times {N_A}\] …………………….Equation 3
Where, m = mass of the unit cell
Volume of a cube = side $ \times $ side $ \times $ side
Let the length of each side of a cube be “a”.
SO, Volume of the cube is ${a^3}$.
V=${a^3}$…………………………………………..equation 4
We know that,
Density of unit cell =$\dfrac{{mass{\text{ }}of{\text{ }}unit{\text{ }}cell}}{{volume{\text{ }}of{\text{ }}unit{\text{ }}cell}}$ $\dfrac{{mass{\text{ }}of{\text{ }}unit{\text{ }}cell}}{{volume{\text{ }}of{\text{ }}unit{\text{ }}cell}}$………………equation 5
On substituting equation 2 and 4 in equation 5 , we arrived at equation 1. This is very important to remember.
According to the question first let’s note down the given values
Given: since it is mentioned it is FCC which is face centered cubic unit cell. One fcc unit cell consists of four atoms therefore, Z= 4.
a = 200pm = $2 \times {10^{ - 8}}$ cm
By using equation 3 , we have to find the molar mass
\[\Rightarrow M = \dfrac{{300}}{{24 \times {{10}^{23}}}} \times 6.022 \times {10^{23}}\]
$= 75.275$ g/mol
We know that, ${N_A}$=$6.022 \times {10^{23}}$
Substituting the value of Z, M, ${N_A}$ and a in equation 1
\[\Rightarrow d = \dfrac{{zM}}{{{a^3}{N_A}}}\]
\[\Rightarrow d = \dfrac{{4 \times 75.275}}{{{{(2 \times {{10}^{ - 8}})}^3} \times 6.022 \times {{10}^{23}}}}\]
\[\Rightarrow d = 62.5g/c{m^3}\]

Thus, the density of the unit cell is $62.5g/c{m^3}$

Note: Always remember the unit of all the quantities:
d = \[g/c{m^3}\]
M = g/mol
${a^3}$= \[c{m^3}\]
The value of a mostly will be given in terms of pm (Picometer) , so we have to make sure we first convert it to centimeter and proceed with the calculation.