The density of crystalline sodium chloride is 2.165 g/$c{{m}^ {-3}} $. What is the edge length of the unit cell? What would be the dimensions of a cube containing one mole of NaCl?

Answer Verified Verified
Hint: The edge means the line segment where the two edges and the cube have 12 edges and the length of one edge to another is called the edge length. It can be calculated by using the density formula i.e. d =$\dfrac{\text{M}\times Z}{{{\text{N}}_{\circ }}\times {{\text{a}}^{3}}\times {{10}^{-23}}}$, where d is the density, ${{\text{a}}^{3}}$ is the edge length,${{\text{N}}_{{}^\circ }}\times {{10}^{-23}}$ is Avogadro number and Z is the no of atoms in a unit cell.

Complete answer:
We can calculate the edge length by using the formula of density of the crystal lattice. The formula of density of a unit cell is as follows:
         Density of unit cell=$\dfrac {\text {mass of the unicell}} {\text {volume of the unicell}} $
Where mass of the unit cell=mass of one atom× no. of atoms in a unit cell
                                          =$\dfrac{mass} {Avogadro\text{no}\text {. (} {{\text{N}} _ {\circ}})} $×Z
So, the density, d=$\dfrac{\text{M}\times Z} {{{\text{N}} _ {\circ}} \times \text {Volume of unit cell}} $
                          d =$\dfrac{\text{M}\times Z}{{{\text{N}}_{\circ }}\times {{\text{a}}^{3}}\times {{10}^{-23}}}$ -----------(1)
Where, ${{\text{a}}^{3}}\times {{10}^{-23}}$ is the volume of the unit cell ,m is the mass of the element and Z is the no of atoms in a unit cell i.e. for bcc=2, fcc=4 and for simple cube Z=1.
In the statement density of the unit cell is given, by putting the value of density in equation (1) we can find out the value of ‘a’.
 Density of crystalline sodium chloride is 2.165 g/$c{{m}^ {-3}} $.
Mass of the element NaCl = 23 +35.5
                                          = 58.5
Since, NaCl occupies fcc arrangement, then Z=4
Put all the values in equation (1), we get:
                    2.165=$\dfrac {58.5\times 4} {6.023\times {{\text{a}} ^ {3}} \times {{10} ^ {-23}}} $
                        ${{\text{a}} ^ {3}} $×=$\dfrac {58.5\times 4} {6.023\times 2.165\times {{10} ^ {-23}}} $
                   ${{\text{a}} ^ {3}} $= 1.794×${{10} ^ {-23}} $
                      a=$\sqrt {1.794\times {{10} ^ {-23}}} $
                      a=5.64×${{10} ^ {-8}} $cm
where a is the edge length.
Now, the edge of cube containing one mole NaCl:
                 ${{\text{a}} ^ {3}} $=$\dfrac {\text {molar mass}} {\text{density}} $
                    a =${{[\dfrac {58.8} {2.165}]} ^ {1/3}} $
Hence, edge length= 3 cm

Keep in mind that the NaCl molecule has fcc arrangement and the no of unit cell (smallest repeating unit in the space lattice which when repeated over and over gives rise to the formation of space lattice) per atom is 4.