Question

Calculate the mobile electron density for nickel. A mole of nickel has a mass of 59g (0.059kg), and one mobile electron is released by each atom in metallic nickel. The density of nickel is about 8.8g/cm³, or$8.8\times {10}^{3}kg/m^3$

Short answer

The mobile electron density is, $8.98\times {10}^{28}kg/m^3$.

Step-by-step solution

Identification of the given data 

The given data can be listed below as,

· The mass of one mole of nickel is, 59g or 0.059 kg.

· The density of nickel is, $\rho =8.8g/c{m}^3$orrole="math" localid="1668551564749" $\rho =8.8\times {10}^{3}kg/m^3$

Significance of the Avogadro’s number

The quantity of particles in one mole of a substance is known as Avogadro's number. It can be used to change moles into particles or particles into particles.

Determination of the mobile electron density for nickel

According to the question,

1 mole of nickel = Avogadro atoms = 0.059kg

$1m^3$of nickel = $\rho kg$

$$\begin{gathered} \rho kg=\frac{\rho }{0.059moles} \\ =Avogadro\times \frac{\rho }{0.059atoms} \end{gathered}$$

The mobile electron density is expressed as,

$$\begin{gathered} =\frac{\rho }{0.059moles} \\ =Avagadro\times \frac{\rho }{0.059atoms} \end{gathered}$$

Substitute all the value in the above equation

The mobile electron density is,

$$\begin{gathered} =6.023\times {10}^{23}\times \frac{8.8\times {10}^{3}kg/m^3}{0.059} \\ =8.98\times {10}^{28}kg/m^3 \end{gathered}$$

Hence the mobile electron density is,$8.98\times {10}^{28}kg/m^3$