91Ó°ÊÓ

a) The gold is monovalent element.

b) Molar mass of gold (Appendix F),A = 197 g/mol

c) Density of gold (Appendix F),$d=19.3\mathrm{g}/{\mathrm{cm}}^3$

Mass of one mole of a sample of element is known as Molar mass. According to Avogadro, one mole of a sample consists of$\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{×}{\mathbf{10}}^{\mathbf{23}}$ atoms, also known as Avogadro’s Number. Thus, if we divide the molar mass with Avogadro’s Number, we will get the mass of one atom of an element.

Formulae:

The mass of an atom, $M=A/N_A,\mathrm{where}{\mathrm{N}}_{\mathrm{A}}=6.022×{10}^{23}{\mathrm{mol}}^{-1}$ (i)

The number density of conduction electrons,$n=\frac{d}{M}$ (ii)

d= density of the atom, M = mass of a single atom

Since each atom contributes one conduction electron, the number of conduction electrons per unit volume should be equal to the number density of atoms of gold present in the sample.

Using equation (i) in equation (ii), the number density of the conduction electrons in gold is given as-

$$\begin{gathered} n=\frac{d}{A/N_A} \\ =\frac{\left(19.3\mathrm{g}/{\mathrm{cm}}^3\right)\left({10}^6{\mathrm{cm}}^3/{\mathrm{m}}^3\right)}{\left(197\mathrm{g}/\mathrm{mol}\right)/\left(6.022×{10}^{23}{\mathrm{mol}}^{-1}\right)} \\ =5.90×{10}^{28}{\mathrm{m}}^3 \end{gathered}$$

Hence, the value of the number density is $5.90×{10}^{28}{\mathrm{m}}^3$.