Chapter 29.1, Problem 1PP

To determine

To calculate: the number of free electrons per centimeter cube in zinc.

Answer to Problem 1PP

The number of free electrons per centimeter cube in zinc is $1.314\times {10}^{23}$ .

Explanation of Solution

Given data:

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

Atomic mass of zinc per mole, $M_a=65.38$ g / mole

Density of zinc, $\rho =7.14$${\text{g/cm}}^3$

Number of electrons per atom = 2

Formula used:

Free electron per cubic centimeter of zinc,

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

Calculation:

Free electron per cubic centimeter of zinc,

  $\begin{array}{l}\frac{\text{Free electron}}{{\text{cm}}^3}=\left(\frac{\text{Free electron}}{1\text{ atom}}\right)\left(N_a\right)\left(\frac{1}{M}\right)\left(\rho \right)\\ \frac{\text{Free electron}}{{\text{cm}}^3}=\left(2\right)\left(6.02\times {10}^{23}\right)\left(\frac{1}{65.38}\right)\left(7.14\right)\\ \frac{\text{Free electron}}{{\text{cm}}^3}=1.314\times {10}^{23}\end{array}$

Conclusion:

Hence, the number of free electrons per centimeter cube in zinc is $1.314\times {10}^{23}$ .