Question

The density and edge length values for a crystallise elemens with $fcc$ lattice are $10g$ $c{m}^{-3}$ and 400 pm, respectively. The number of unit cell in $32g$ of this crystal is:

• A. $8\times {10}^{23}$
• B. $5\times {10}^{22}$
• C. $8\times {10}^{22}$
• D. $5\times {10}^{23}$

Answer 1

The correct option is B $5\times {10}^{22}$

Theoretical density of crystal, $\rho =\frac{nM}{N_o{a}^3}g/c{m}^3$

Given that for fcc, $n=4,a=400\times {10}^{-10}cm,N_o=6\times {10}^{23}\rho =10g/c{m}^3$

$\therefore M=\frac{\rho .N_o.a^3}{n}$

$=\frac{10\times 6\times {10}^{23}\times (400{)}^3\times {10}^{-30}}{2}$

$\Rightarrow M=96.4g$

Number of moles in $32g$ of crystal $=\frac{32}{96.4}=0.33$

Number of atoms $=0.33\times 6.022\times {10}^{23}$

Now, in fcc, there are $4$ atoms in a unit cell.

So, number of unit cells $=\frac{0.33}{4}\times 6.022\times {10}^{23}=5\times {10}^{22}$