Chapter 29.1, Problem 2PP

To determine

The number of free electrons in 1 ${\text{cm}}^3$ of silver atom.

Answer to Problem 2PP

The number of free electrons per centimeter cube in silver is $5.86\times {10}^{22}$ .

Explanation of Solution

Given data:

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

Atomic mass of silver, $M=107.9$ g

Density of zinc, $\rho =10.5$${\text{g/cm}}^3$

Given number of free electron per atom = 1

Formula used:

Free electrons per centimeter cube,

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

Calculation:

Free electrons per centimeter cube,

  $\begin{array}{l}\frac{\text{Free electron}}{{\text{cm}}^3}=\left(\frac{\text{Free electron}}{\text{atom}}\right)\left(N_a\right)\left(\frac{1}{M}\right)\left(\rho \right)\\ \frac{\text{Free electron}}{{\text{cm}}^3}=\left(1\right)\left(6.02\times {10}^{23}\right)\left(\frac{1}{107.9}\right)\left(10.5\right)\\ \frac{\text{Free electron}}{{\text{cm}}^3}=5.86\times {10}^{22}\end{array}$

Conclusion:

Hence, the number of free electrons per centimeter cube in silver is 5.86 x 10²².