Chapter 29.1, Problem 3PP

To determine

The number of free electrons in $1{\text{ cm}}^3$ of gold atom by using the reference tables.

Answer to Problem 3PP

The number of free electrons per centimeter cube in gold is $5.90\times {10}^{23}$ .

Explanation of Solution

Given data:

Number of atoms per mole in gold, $N_a=6.02\times {10}^{23}$ atoms / mole

Atomic mass of gold, $M=197$ g

Density of gold, $\rho =19.31$${\text{g/cm}}^3$

1 atom of the gold contains only 1 free electron

Formula used:

Number of free electrons in $1{\text{ cm}}^3$ of gold,

  $\frac{\text{Free electron}}{{\text{cm}}^3}=\left(\frac{\text{Free electron}}{\text{1 atom}}\right)\left(N_a\right)\left(\frac{1}{M}\right)\left(\rho \right)$

Calculation:

Number of free electrons in $1{\text{ cm}}^3$ of gold,

  $\begin{array}{l}\frac{\text{Free electron}}{{\text{cm}}^3}=\left(\frac{\text{Free electron}}{\text{1 atom}}\right)\left(N_a\right)\left(\frac{1}{M}\right)\left(\rho \right)\\ \frac{\text{Free electron}}{{\text{cm}}^3}=\left(1\right)\left(6.02\times {10}^{23}\right)\left(\frac{1}{197}\right)\left(19.31\right)\\ \frac{\text{Free electron}}{{\text{cm}}^3}=5.90\times {10}^{22}\\ \end{array}$

Conclusion:

The number of free electrons per centimeter cube in gold is $5.90\times {10}^{23}$ .