When air pollution is high, ozone \(\left(\mathrm{O}_{3}\right)\) contents can reach \(0.60 \mathrm{ppm}\) (i.e., \(0.60\) mol ozone per million mol air). How many molecules of ozone are present per liter of polluted air if the barometric pressure is \(755 \mathrm{~mm} \mathrm{Hg}\) and the temperature is \(79^{\circ} \mathrm{F} ?\)

Short Answer

Expert verified
There are approximately \(1.468 \times 10^{19}\) ozone molecules present in one liter of polluted air under the given conditions.

Step by step solution

01

Convert temperature and pressure to SI units

First, we need to convert the given temperature from Fahrenheit to Celsius and then to Kelvin, and the given pressure from mm Hg to atm. Temperature: $$ T(^{\circ}C) = \frac{5}{9} \left(T(^{\circ}F) - 32 \right) = \frac{5}{9} \left(79 - 32 \right) = 26.11^{\circ}C $$ $$ T(K) = T(^{\circ}C) + 273.15 = 26.11 + 273.15 = 299.26 K $$ Pressure: $$ P(atm) = \frac{P(mmHg)}{760} = \frac{755}{760} = 0.9934 atm $$
02

Calculate the number of moles of air in 1 L of polluted air

Now, we'll use the Ideal Gas Law to find the amount of air in 1L of polluted air: $$ PV = nRT $$ where, P = Pressure in atm (0.9934 atm) V = Volume in L (1 L) R = Ideal Gas constant, 0.0821 L.atm/K.mol T = Temperature in K (299.26 K) n = Number of moles (our goal) We need to solve for n: $$ n = \frac{PV}{RT} = \frac{(0.9934)(1)}{(0.0821)(299.26)} = 0.04065 mol $$
03

Calculate the number of moles of ozone in 1 L of polluted air

We are given that the ozone concentration in the air is 0.60 ppm. Therefore, ozone makes up 0.60 mol per million moles of air. We can use this information to find the number of moles of ozone in 1 L of polluted air. $$ \frac{0.60 \ mol \ ozone}{10^6 \ mol \ air} = \frac{x \ mol \ ozone}{0.04065 \ mol \ air} $$ Now, solve for x: $$ x = \frac{0.60 \ mol \ ozone \times 0.04065 \ mol \ air}{10^6 \ mol \ air} = 2.439 \times 10^{-5} \ mol \ ozone $$
04

Calculate the number of molecules of ozone

Finally, we can use Avogadro's number to convert moles of ozone into molecules of ozone: $$ \text{Number of molecules} = \text{moles} \times \text{Avogadro's number} = 2.439 \times 10^{-5} \ mol \ ozone \times 6.022 \times 10^{23} \ molecules/mol = 1.468 \times 10^{19} \ molecules $$ So, there are approximately \(1.468 \times 10^{19}\) molecules of ozone per liter of polluted air under the given conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry and physics that describes the behavior of gases under various conditions. It is expressed by the equation
\[PV = nRT\]
where P represents the pressure of the gas, V is the volume it occupies, n is the amount in moles, R is the ideal gas constant (0.0821 L.atm/K.mol), and T is the temperature in Kelvin (K). This law allows us to calculate any one of these variables if the others are known.
When solving problems related to the concentration of substances in the air, such as the given exercise, the Ideal Gas Law is indispensable. It enables us to determine the volume of air—or any gas—in a given space, which is essential for calculating concentrations.

Application in Pollution Measurement

In environmental studies, the Ideal Gas Law helps assess the amount of pollutants in the air by establishing the relationship between gas volumes and mole quantities. By figuring out the number of moles of a gas in a specific volume, we can proceed to find the concentration of the pollutants, provided that their volumetric proportions are known.
Avogadro's Number
Avogadro's number is a constant that is central to chemistry and defines the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. Avogadro's number is approximately
\(6.022 \times 10^{23}\)
particles per mole.
This number makes a bridge between the macroscopic world we can measure and the microscopic world of molecules and atoms. So, when we talk about a 'mole' of any substance, we are referring to a quantity that contains as many elementary entities as there are atoms in exactly 12 grams of carbon-12, which is Avogadro's number.

Connecting Moles to Molecules

Using Avogadro's number, we can convert moles of a gas into individual molecules, which is exactly what was done in the solution to determine the number of ozone molecules per liter of air. This conversion is essential in environmental chemistry, where it's often necessary to discuss quantities at the molecular level to understand the impact of pollutants on the environment and human health.
Environmental Chemistry
Environmental chemistry is the branch of science that focuses on the chemical phenomena occurring in the environment. It involves the study of the sources, reactions, transport, effects, and fates of chemical species in the water, soil, and air environments, as well as the effect of human activity upon these.
Understanding environmental chemistry is essential when addressing issues like air pollution. Pollutants like ozone (\(\text{O}_3\)) can have significant health impacts, and their concentration in the atmosphere needs to be carefully monitored. The concentration of ozone and other pollutants in the air is often expressed in parts per million (ppm), as seen in the textbook example.

Role in Pollution Tracking

Environmental chemistry provides the tools and concepts needed to quantify and analyze pollutants. By applying principles such as the Ideal Gas Law and using Avogadro's number, we can calculate the exact amount of a given substance in a particular volume of air, enabling us to better understand and mitigate potential environmental risks and ensure regulatory compliance.

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Most popular questions from this chapter

Cyclopropane mixed in the proper ratio with oxygen can be used as an anesthetic. At \(755 \mathrm{~mm} \mathrm{Hg}\) and \(25^{\circ} \mathrm{C}\), it has a density of \(1.71 \mathrm{~g} / \mathrm{L}\). (a) What is the molar mass of cyclopropane? (b) Cyclopropane is made up of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\). What is the molecular formula of cyclopropane?

The contents of a tank of natural gas at \(1.20 \mathrm{~atm}\) is analyzed. The analysis showed the following mole percents: \(88.6 \% \mathrm{CH}_{4}, 8.9 \% \mathrm{C}_{2} \mathrm{H}_{6}\), and \(2.5 \%\) \(\mathrm{C}_{3} \mathrm{H}_{3}\). What is the partial pressure of each gas in the tank?

Dichlorine oxide is used as bactericide to purify water. It is produced by the chlorination of sulfur dioxide gas. $$ \mathrm{SO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SOCl}_{2}(l)+\mathrm{Cl}_{2} \mathrm{O}(g) $$ How many liters of \(\mathrm{Cl}_{2} \mathrm{O}\) can be produced by mixing \(5.85 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) and \(9.00 \mathrm{~L}\) of \(\mathrm{Cl}_{2}\) ? How many liters of the reactant in excess are present after reaction is complete? Assume \(100 \%\) yield and that all the gases are measured at the same temperature and pressure.

Some chambers used to grow bacteria that thrive on \(\mathrm{CO}_{2}\) have a gas mixture consisting of \(95.0 \% \mathrm{CO}_{2}\) and \(5.0 \% \mathrm{O}_{2}\) (mole percent). What is the partial pressure of each gas if the total pressure is \(735 \mathrm{~mm} \mathrm{Hg}\) ?

At \(25^{\circ} \mathrm{C}\) and \(380 \mathrm{~mm} \mathrm{Hg}\), the density of sulfur dioxide is \(1.31 \mathrm{~g} / \mathrm{L}\). The rate of effusion of sulfur dioxide through an orifice is \(4.48 \mathrm{~mL} / \mathrm{s}\). What is the density of a sample of gas that effuses through an identical orifice at the rate of \(6.78 \mathrm{~mL} / \mathrm{s}\) under the same conditions? What is the molar mass of the gas?

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