Question

A compound $\text{CuCl}$ has face-centred cubic structure. Its density is $3.05\text{g}{\text{cm}}^{-3}$. The length of unit cell is:
(Given, $N_A=6\times {10}^{23}$)

• A. $12\stackrel{\circ }{\text{A}}$
• B. $8\stackrel{\circ }{\text{A}}$
• C. $9\stackrel{\circ }{\text{A}}$
  
• D. $6\stackrel{\circ }{\text{A}}$
  

Answer 1

The correct option is D $6\stackrel{\circ }{\text{A}}$

Let, the length of the cell be $a$

We Know

Density$=\frac{Z\times M}{V\times N_A}...\left(1\right)$

Given that

$\text{Density}=3.05\text{g}{\text{cm}}^{-3}$
${\text{N}}_{\text{A}}=6\times {10}^{23}$
For, face-centred cubic unit cell, $Z=4$
Molecular Mass of $\text{CuCl}=99\text{u}$
Putting all the values in equation $\left(1\right)$, we get
$3.05=\frac{4\times 99}{a^3\times 6\times {10}^{23}}\Rightarrow a^3=\frac{4\times 99}{3.05\times 6\times {10}^{23}}\Rightarrow a^3=216.39\times {10}^{-24}{\text{cm}}^3\Rightarrow a\approx 6\stackrel{\circ }{\text{A}}$
The length of unit cell is $6\stackrel{\circ }{\text{A}}$