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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}} $

a=3cm

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.