Question

Question: One form of crystalline iron has a bcc lattice with an iron atom at every lattice point. Its density at $\mathbf{\text{25}}\mathbf{°}\mathbf{\text{C}}$is ${\mathbf{\text{7.86g cm}}}^{\mathbf{-}\mathbf{3}}$. The length of the edge of the cubic unit cell is role="math" localid="1660653062517" $\mathbf{2}\mathbf{.}\mathbf{87}\stackrel{\mathbf{\circ }}{\mathbf{A}}$. Use these facts to estimate Avogadro’s number.

Short answer

The estimated Avogadro’s number is $6.0113\times {10}^{23}/mol$.

Step-by-step solution

The volume of this cell in cubic angstroms.

Given;

The length of the edge of the cubic unit cell is $2.87\stackrel{\circ }{\mathrm{A}}$.

Therefore, the volume becomes;

$$\begin{gathered} {\text{a}}^3=2.87\stackrel{\circ }{A}\times 2.87\stackrel{\circ }{A}\times 2.87\stackrel{\circ }{A} \\ {\text{a}}^3=23.64\stackrel{\circ }{A^3} \end{gathered}$$

Changing Angstrom to cm;

$$\begin{gathered} 23.64\stackrel{\circ }{A^3}=23.64\times {\left({\text{10}}^{-8}\right)}^{3}c{m}^3 \\ 23.64\stackrel{\circ }{A^3}=23.64\times {\left({\text{10}}^{-8}\right)}^{3}c{m}^3 \end{gathered}$$

The estimated Avogadro’s number.

As known;

$density=\frac{M}{volume\times N_A}$

$density=\frac{z\times M}{volume\times N_A}$

For bcc; z = 2

Atomic mass of Fe is 55.845.

For 2 atoms in one unit cell; molecular mass becomes;

$\begin{array}{rcl}z\times M& =& 55.845\times 2\\ & =& 111.69\end{array}$

Putting the values in the above equation;

$7.86{\text{g cm}}^{-3}=\frac{111.69g/mol}{23.64\times {\text{10}}^{-24}c{m}^3\times N_A(/mol)}$

localid="1660653822848" $$\begin{gathered} N_A(/mol)=\frac{111.69g/mol}{7.86{\text{g cm}}^{-3}\times 23.64\times {\text{10}}^{-24}c{m}^3} \\ {N}_A(/mol)=\frac{111.69g/mol}{185.8\times {\text{10}}^{-24}} \\ {N}_A(/mol)=\frac{111.69/mol}{185.8\times {\text{10}}^{-24}} \\ {N}_A(/mol)=6.0113\times {10}^{23}/mol \end{gathered}$$