Question

The density of argon (face centered cubic cell) is $1.83g/c{m^3}$ at ${20^ \circ }C$. What is the length of an edge in a unit cell?
A.$0.599nm$
B.$0.569nm$
C.$0.525nm$
D.$0.551nm$

Answer 1

Hint: We have to calculate the numbers of atoms present in the unit cell of FCC atom and then, we have to calculate the mass of the unit cell of FCC type using molar mass of argon and Avogadro number. From the density and mass of the unit cell, we have to calculate the volume of the unit cell. From the volume of the unit cell, we can calculate the edge of the unit cell.

Complete step by step solution:
Given data contains,
Argon is a face centered cube
Density of argon is $1.83g/c{m^3}$.
Molar mass of argon is $40g/mol$.
First, we have to calculate the number of atoms present in the unit cell of FCC.
Number of atoms=$\dfrac{1}{8} \times 8 + 6 \times \dfrac{1}{2}$
Number of atoms=$4$ atoms
The number of atoms present in the unit cell of FCC is 4 atoms.
Next, we will calculate the mass of the unit cell of FCC using the molar mass of argon and Avogadro number.
The value of Avogadro number is $6.022 \times {10^{23}}$.
Mass of unit cell=$4 \times \dfrac{{40}}{{6.022 \times {{10}^{23}}}}$
Mass of unit cell=$2.65 \times {10^{ - 22}}g$
Mass of unit cell=$2.65 \times {10^{ - 22}}g$
The mass of the unit cell is $2.65 \times {10^{ - 22}}kg$.
We will now calculate the volume of the unit cell using the density of the unit cell.
We know the formula of density as,
Density of the unit cell$ = \dfrac{{{\text{Mass of unit cell}}}}{{{\text{Volume of unit cell}}}}$
Let us substitute the values of the mass and density of the unit cell
$1.83$=$\dfrac{{2.65 \times {{10}^{ - 22}}}}{{{a^3}}}$
On simplifying we get,
$ \Rightarrow {a^3} = 1.44 \times {10^{ - 22}}c{m^3}$
We shall take the cube root of the volume of the unit cell
${a^3} = 1.44 \times {10^{ - 22}}$
$a = \sqrt[3]{{1.44 \times {{10}^{ - 24}}}}cm$
$ \Rightarrow a = 5.25 \times {10^{ - 8}}cm$
The volume of the unit cell that is the edge length of the unit cell is $5.25 \times {10^{ - 8}}cm$.
The length in centimeter is converted into nanometer as,
$nm = 5.25 \times {10^{ - 8}}cm \times \dfrac{{{{10}^7}nm}}{{1cm}}$
$nm = 0.525nm$
The length of an edge of a unit cell is $0.522nm$.
Therefore,option (C) is correct.

Note:
Now we can see an example of a compound that has face-centered cubic lattice is sodium chloride. Lithium fluoride, lithium chloride, Sodium fluoride, potassium fluoride, potassium chloride etc are examples of compounds that contain face centered cubic structures. Simple cubic cell, and body centered cubic cell are the other two crystal structures.