Question

An element with molar mass 64 gmol -1 and density 6.6 g cm -3 forms a cubic unit cell.The edge length of unit cell is 4 x 10 -8cm.The type of cubic unit cell formed is ?

Answer 1

Answer:

FCC is the answer

please mark me brilliant

Answer 2

Answer:

The type of the cubic unit cell formed with Z = $4$ is fcc ( face-centered cubic ).

Explanation:

Given,

The molar mass of the element, M = $64g/mol$.

The density of the given element, d = $6.6g/cm^{3}$

The edge length of the given element, a = $4\times 10^{-8}cm$

The type of the cubic unit cell formed is =?

As we know,

• The number of atoms present per unit cell ( Z ) is fixed for cubic unit cells.

Like,

• For bcc ( body-centered cubic ) Z = 2

• For fcc ( face-centered cubic ) Z = 4

Now, we have to calculate the value of Z.

As we know,

• density = $\frac{Z\times M}{N_{A} \times a^{3} }$

Here, $N_{A}$ = Avogadro number = $6.022\times 10^{23} units$

Thus,

• Z = $\frac{d \times N_{A}\times a^{3} }{M}$

After putting the values in the equation, we get:

• Z = $\frac{6.6 \times 6.022\times 10^{23} \times (4\times 10^{-8}) ^{3} }{64}$
• Z = $3.975$ ≈ $4$

As we know, for fcc ( face-centered cubic ) Z = 4

Hence, the type of cubic unit cell formed is fcc ( face-centered cubic ).