A sample of a certain metal has a volume of $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{m}}^{\mathbf{3}}$. The metal has a density of $\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$and a molar mass of 60 g/mol. The atoms are bivalent. How many conduction electrons (or valence electrons) are in the sample?

a) Volume of the sample,$V=4\times {10}^{-5}\hspace{0.33em}{m}^3$

b) Density of the metal,$d=9\hspace{0.33em}g/c{m}^3$

c) Molar mass of the metal,$A=60\hspace{0.33em}g/mol$

d) The atoms are bivalent.

In metallic elements, the electrons present in the valence band absorb energy from the surrounding and jump from the valence band to the conduction band. Now these electrons are not bound to any atom and are free to move within the walls of the sample. These electrons are known as conduction electrons. The blank space that is left behind in the valence band are called holes. Both the conduction electrons and the holes are responsible for the current in the conductor.

Formula:

The mass of a substance using the density,$m=d\times V$ (i)

The number of moles of a substance,$n=\frac{m}{A}$ (ii)

where, A is the molar mass of the substance

The total number of atoms present in the sample,

$N=n{N}_A$ (iii)

Where $N_A=6.022\times {10}^{23}/mol$is called Avogadro’s number.

Using the given data in equation (i), the mass of the sample is calculated as follows:

$$\begin{gathered} m=\left(9g/c{m}^3\right)\left(40c{m}^3\right) \\ =360\mathrm{g} \end{gathered}$$

Now, using this mass value in equation (ii), we can calculate the number of moles present in the sample-

$$\begin{gathered} n=\frac{360g}{60g/mol} \\ =6.0\mathrm{mol} \end{gathered}$$

But, we are given that the atoms are bivalent, that means each atom contribute 2 electrons each. Thus, the total conduction electrons in the sample are.

Thus, the number of conductions electrons in the sample is given using equation (iii) as follows:

$$\begin{gathered} N=\left(12mol\right)\left(6.022\times {10}^{23}/mol\right) \\ =7.2\times {10}^{24} \end{gathered}$$

Hence, the number of the conduction electrons in the sample is $7.2\times {10}^{24}$.