Q12.129CP - Chemistry: Molecular Nature Of Matter And Change

Question

Iron crystallizes in a body-centered cubic structure. The volume of one Fe atom is $\mathbf{8}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{-24}{\mathbf{cm}}^3$, and the density of Fe is  $\mathbf{7}\mathbf{.}\mathbf{874}{\mathbf{gcm}}^{-3}$ . Find an approximate value for Avogadro’s number.

Step-by-Step Solution

Answer

The approximate value for Avogadro’s number is $\mathbf{5}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$   atoms/molecules having a body-centered cubic structure.

Step 1: Definition

Density can be defined as the ratio of mass to volume. Density is inversely proportional to the volume of the unit cell as the mass of the atom is constant.

 

Density equation for the unit cell:

 $\mathbf{\rho }\mathbf{=}\frac{Z\times M}{\mathbf{Na}\mathbf{\times }{\mathbf{a}}^3}$

 

Here,

ρ= Density,  

Z= Number of atoms in contact in a particular structure

M = Molar mass 

${\mathrm{N}}_{\mathrm{a}}$ = Avogadro number

a = atomic radius.

 

For BCC, the value of  $\text{Z = 2}$  which is the number of atoms in contact in a particular structure.

Step 2: The radius (r) of the atom in the cubic lattice.

The volume of the body-centered structure =  ${\text{8.38×10}}^{\text{-24}}{\text{cm}}^{\text{3}}$.

 $\begin{array}{c}\mathbf{V}\mathbf{=}\frac{4}{3}{\mathbf{\pi r}}^3\\ \mathbf{8}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{-24}\mathbf{=}\frac{4}{3}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{14}\mathbf{\times }{\mathbf{r}}^3\\ {\mathbf{r}}^3\mathbf{=}\frac{3}{4}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{14}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{-24}\\ \mathbf{r}\mathbf{=}{\{\mathbf{2}\mathbf{\times }{\mathbf{10}}^{-24}\}}^{\frac{1}{3}}\\ \mathbf{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{-8}\mathbf{cm}\end{array}$

 

 

The radius of the atom in the cubic lattice $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{-8}\mathbf{cm}$ .

Step 3: The length (a) of the side body-centred cubic lattice.

$\begin{array}{c}\mathrm{a}=\frac{4}{\sqrt{3}}\mathrm{r}\\ \mathrm{a}=\frac{4}{\sqrt{3}}\times 1.26\times {10}^{-8}\\ \mathrm{a}=\frac{4}{1.732}\times 1.26\times {10}^{-8}\\ \mathrm{a}=3\times {10}^{-8}\mathrm{cm}\end{array}$

Step 4: The approximate value for Avogadro’s number.

Atomic radius, $\mathbf{a}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{-8}$ 

Density of the unit cell,  $\mathbf{\rho }\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{874}{\mathbf{gcm}}^{-3}$

Molar mass of atom or molecule,  $\text{m = 56 g/mole}$

 $\begin{array}{c}\mathbf{\rho }\mathbf{=}\frac{Z\times M}{\mathbf{Na}\mathbf{\times }{\mathbf{a}}^3}\\ \mathbf{7}\mathbf{.}\mathbf{874}\mathbf{=}\frac{2\times 56}{\mathbf{NA}\mathbf{\times }{(\mathbf{3}\mathbf{\times }{\mathbf{10}}^{-8})}^3}\\ \mathbf{NA}\mathbf{=}\frac{112}{\mathbf{7}\mathbf{.}\mathbf{874}\mathbf{\times }\mathbf{27}\mathbf{\times }{\mathbf{10}}^{-24}}\\ \mathbf{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{24}\\ \mathbf{a}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{23}\mathbf{atoms}\mathbf{/}\mathbf{molecules}\end{array}$

 

 

The approximate value for Avogadro’s number is $\mathbf{5}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$  atoms/molecules.