Chapter 29, Problem 75A

To determine

To design a problem with real-life objects, the solution of which would include the given expression.

Answer to Problem 75A

To find the number of free electrons per cubic centimeter of cadmium, whose density is $\text{8}{\text{.645 g/cm}}^3$ and its atomic mas is $112.411\text{ g/mol}$ , having $2$ free electrons per atom the given expression can be used.

Explanation of Solution

Given info:

The expression:

  $\left(\frac{2e^{-1}}{1\text{ atom }}\right)\left(\frac{6.02\times {10}^{23}\text{ atoms }}{1\text{ mole}}\right)\left(\frac{1\text{ mole}}{112.411\text{ g}}\right)\left(\frac{8.645\text{ g}}{{\text{cm}}^3}\right)$

Calculation:

As per problem

The given expression is,

  $\left(\frac{2e^{-1}}{1\text{ atom }}\right)\left(\frac{6.02\times {10}^{23}\text{ atoms }}{1\text{ mole}}\right)\left(\frac{1\text{ mole}}{112.411\text{ g}}\right)\left(\frac{8.645\text{ g}}{{\text{cm}}^3}\right)$

It is use to find the number of free electrons in a cubic centimeter of an element and is written generally in the form

  $\text{Free electrons per cubic centimeter = }\frac{\left(\text{number of atoms per mole}\right)\left(\text{density of element}\right)}{\left(\text{atomic mass}\right)}$

From the provided expression,

Number of free electrons per atom= $2$

Number of atoms per mole= $6.02\times {10}^{23}$

Atomic mass of the element = $112.411\text{ g/mol}$

Density of the element= $\text{8}{\text{.645 g/cm}}^3$

An element with atomic mass $112.411\text{ g/mol}$ and density $\text{8}{\text{.645 g/cm}}^3$ is cadmium.

To find the number of free electrons per cubic centimeter of cadmium, whose density is $\text{8}{\text{.645 g/cm}}^3$ and its atomic mas is $112.411\text{ g/mol}$ , having $2$ free electrons per atom the given expression can be used.

Free electrons per cubic centimeter of cadmium= $\left(\frac{2e^{-1}}{1\text{ atom }}\right)\left(\frac{6.02\times {10}^{23}\text{ atoms }}{1\text{ mole}}\right)\left(\frac{1\text{ mole}}{112.411\text{ g}}\right)\left(\frac{8.645\text{ g}}{{\text{cm}}^3}\right)$

Conclusion:

Thus, to find the number of free electrons per cubic centimeter of cadmium, whose density is $\text{8}{\text{.645 g/cm}}^3$ and its atomic mas is $112.411\text{ g/mol}$ , having $2$ free electrons per atom the given expression can be used.