Chapter 21, Problem 21QP

Interpretation Introduction

Interpretation:

The number of ozone molecules in the stratosphere and their mass in kg are to becalculated.

Concept introduction:

Stratosphere is the second layer of the Earth’s atmosphere and contains oxygen, ozone,and nitrogen gases.

Ozone is an allotrope of oxygen, containing three oxygen atoms. Ozone is formed in the stratosphere.

Answer to Problem 21QP

Solution: The number of ozone molecule is $4.0\times {10}^{37}$ and the mass of ${\text{O}}_3$ is $3.2\times {10}^{12}\text{ kg}$

Explanation of Solution

Given information: The quantity of ozone in the stratosphere is equivalent to a $3.0\text{ mm}$ thick layer of ozone on Earth at STP, the radius of Earth is 6371 km and the surface area of a sphere is $4\pi r^2$.

The number of ozone molecules in the stratosphere can be calculated with the help of volume.

Area of Earth surface=$4\pi r^2$

=$4\times 3.14\times {\left(6371\times {10}^3\text{m}\right)}^2$

=$5.098\times {10}^{14}{\text{m}}^2$

The volume of ozone layer is given by the following expression:

$V=4\pi r^{2}h$

Here, $r$ is the radius of the sphere $\left(6.371\times {10}^3\text{ m}\right)$ and $h$ is the height of the sphere $\left(3.0\text{ mm = 3}\text{.0}\times {\text{10}}^{-3}\text{ m}\right)$

Now substitute the values in the above equation as shown below:

$\begin{array}{l}V=4\pi {\left(6.371\times {10}^3\text{ m}\right)}^2\left(3.0\times {10}^{-3}\text{ m}\right)\\ V=\left(5.098\times {10}^{14}{\text{m}}^2\right)\left(3.0\times {10}^{-3}\text{ m}\right)\\ V=1.5\times {10}^{12}\cancel{{\text{m}}^3}\times \frac{1000\text{ L}}{{10}^{-3}\cancel{{\text{m}}^3}}\\ =1.5\times {10}^{15}\text{L}\end{array}$

$V=1.5\times {10}^{15}\text{ L}$

Therefore, the volume of thesphereis $1.5\times {10}^{15}\text{ L}$.

The number of moles of ozone $\left({\text{O}}_3\right)$ is calculated as follows:

Moles of ${\text{O}}_3$= volume of sphere $\times$ one mole of gas

At standard temperature and pressure (STP), one mole of gas occupies $22.41\text{ L}$

By substituting the values in the above equation:

Moles of ${\text{O}}_3$ = $\left(1.5\times {10}^{15}\cancel{\text{L}}\right)\left(\frac{1\text{ mol}}{22.41\cancel{\text{L}}}\right)$

Moles of ${\text{O}}_3$ =$6.7\times {10}^{13}{\text{ mol O}}_3$

Therefore, the number of moles of ${\text{O}}_3$ is $6.7\times {10}^{13}\text{ mole}$.

The molecule of ozone $\left({\text{O}}_3\right)$ is calculated as follows:

A molecule of ${\text{O}}_3$=moles of ozone $\times$ Avogadro number

By substituting the values in the above equation:

Molecule of ${\text{O}}_3$= $\begin{array}{l}\left(6.7\times {10}^{13}{\text{ mol O}}_3\right)\frac{6.022\times {10}^{23}\text{ molecule}}{1\text{ mol}}\\ \end{array}$

Molecule of ${\text{O}}_3$= $4.0\times {10}^{37}\text{ molecules}$

Therefore, the number ofozonemolecules is $4.0\times {10}^{37}$.

The mass of ozone $\left({\text{O}}_3\right)$ is calculated as follows:

Mass of ${\text{O}}_3$= moles of ozone $\times$ molar mass of ozone

Mass of ${\text{O}}_3$= $$\begin{array}{l}\left(6.7\times {10}^{13}\cancel{{\text{mol O}}_3}\right)\times \frac{\text{48}\text{.00 }\cancel{\text{g}}{\text{ O}}_3}{1\cancel{{\text{mol O}}_3}}\times \frac{1\text{ kg}}{1000\cancel{\text{g}}}\\ \end{array}$$

Mass of ${\text{O}}_3$= $3.2\times {10}^{12}\text{ kg}$

Conclusion

The number of ozone molecules in the stratosphere and their mass in kg are $4.0\times {10}^{37}\text{ molecules}$ and $3.2\times {10}^{12}\text{ kg}$,respectively.