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Chapter 13: Problem 60

Breathing air that contains 4.0\(\%\) by volume \(\mathrm{CO}_{2}\) over time causes rapid breathing, throbbing headache, and nausea, among other symptoms. What is the concentration of \(\mathrm{CO}_{2}\) in such air in terms of (a) mol percentage, (b) molarity, assuming 1 atm pressure and a body temperature of \(37^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The concentration of CO2 in the air is (a) 4.0% mol percentage, and (b) 0.0016 mol/L in terms of molarity, assuming 1 atm pressure and a body temperature of 37掳C.

Step by step solution

01

Convert volume percentage to mol percentage

The given volume percentage of CO2 is 4.0%. In a mixture of gases, the volume percentage is equal to the mol percentage. Hence, mol percentage of CO2 is 4.0%.
02

Find the molarity of CO2

We will use the Ideal Gas Law formula: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Given: - Pressure (P) = 1 atm - Temperature (T) = 37掳C = 310.15 K (converting to Kelvin) - Volume percentage of CO2 = 4.0% First, we need to find the partial pressure of CO2 in the air using the given volume percentage of CO2. Partial pressure of CO2 = total pressure 脳 volume percentage of CO2: \[P_{\mathrm{CO}_2} = 1\,\mathrm{atm} \times 0.04\] \[P_{\mathrm{CO}_2} = 0.04\,\mathrm{atm}\] Now, we will use the Ideal Gas Law formula to find the molarity of CO2. Rearranging the formula for molarity (n/V): \[\frac{n}{V} = \frac{P}{RT}\] Substitute the values: \[\frac{n}{V} = \frac{0.04\,\mathrm{atm}}{(0.0821\,\mathrm{L\,atm/(mol\,K)})\times 310.15\,\mathrm{K}}\] Now, calculate the molarity: \[\frac{n}{V} = 0.0016\,\mathrm{mol/L}\] So, the concentration of CO2 in terms of (a) mol percentage is 4.0%, and (b) molarity is 0.0016 mol/L.

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Key Concepts

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

Volume Percentage
Volume percentage in the context of gases is a way of expressing the concentration of a particular component in a mixture. In this case, it means what portion of the total volume of air is composed of carbon dioxide (CO2). Imagine a balloon filled with air where 4% of the space inside is occupied by CO2 molecules. The rest of the 96% is made up of other gases like nitrogen and oxygen.

Volume percentage becomes especially relevant in gases due to their nature of freely mixing and occupying space uniformly. Thus, saying that air contains 4% CO2 by volume directly tells us about its concentration and potentially its danger level since higher concentrations can affect health by causing issues such as nausea and headaches. This kind of direct percentage is helpful for easy understanding and comparison.
Mol Percentage
Mol percentage, often called mole percent, is very similar to volume percentage when dealing with gases. This is because, under the same conditions of temperature and pressure, gases have volumes directly proportional to the number of moles they have due to Avogadro's law. Therefore, the volume percentage of a gas in a mixture also equals its mol percentage.

In the exercise, the volume percentage of CO2, which is 4%, directly translates to its mol percentage. This shift from volume to mol perspective can be incredibly useful in calculations involving chemical reactions, where mole balance or stoichiometry needs to be considered as it provides a direct link through Avogadro鈥檚 law.
Molarity
Molarity is another way to measure concentration, but unlike volume percentage and mol percentage, it describes the number of moles of a substance in a given volume of solution, typically in liters. It is a commonly used unit in chemistry to express concentrations in solutions. However, for gases, it requires some calculations as gases do not naturally come in liquid solutions.

To find the molarity of CO2 in air, we use the Ideal Gas Law, which relates pressure, volume, and temperature to the number of moles (\(PV = nRT\)). To make it into molarity (\(\frac{n}{V}\)), divide the number of moles by the volume. Using the known conditions of the environment, like atmospheric pressure and body temperature, it can be calculated. Here in the step-by-step solution, we found it to be 0.0016 mol/L for CO2.
Ideal Gas Law
The Ideal Gas Law is a crucial equation in chemistry that relates four important state properties of gases: pressure (P), volume (V), temperature (T), and the number of moles (n). The relationship is given by the equation \(PV = nRT\), where \(R\) is the ideal gas constant. This equation helps us understand how gases will behave under different conditions.

In solving our exercise, we apply the Ideal Gas Law to determine the molarity of CO2 in air. By knowing other conditions, such as pressure set at 1 atm and temperature at 37掳C (310.15 K), we can rearrange the formula to solve for molarity: \(\frac{n}{V} = \frac{P}{RT}\). This equation is powerful because it allows for the conversion between physical conditions and chemical amounts, enabling us to calculate concentrations even for gaseous substances. In this specific case, it highlights how a change in conditions or gas concentration would affect its measurable quantities.

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Most popular questions from this chapter

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right),\) the solubility of \(\mathrm{N}_{2}\) in water at ordinary atmospheric pressure \((1.0 \mathrm{atm})\) is 0.015 \(\mathrm{g} / \mathrm{L} .\) Air is approximately 78 \(\mathrm{mol} \% \mathrm{N}_{2}\) . (a) Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, assuming blood is a simple aqueous solution. (b) At a depth of 100 \(\mathrm{ft}\) in water, the external pressure is 4.0 atm. What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? (c) If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

(a) A sample of hydrogen gas is generated in a closed container by reacting 2.050 g of zinc metal with 15.0 \(\mathrm{mL}\) . of 1.00 \(\mathrm{M}\) sulfuric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is 122 mL. Calculate the partial pressure of the hydrogen gas in this volume at \(25^{\circ} \mathrm{C}\) , ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is \(7.8 \times 10^{-4} \mathrm{mol} / \mathrm{L}\) -atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)?

Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of 0.64 g of adrenaline in 36.0 g of \(\mathrm{CCl}_{4}\) elevates the boiling point by \(0.49^{\circ} \mathrm{C}\) Calculate the approximate molar mass of adrenaline from this data.

Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called "salting out" the protein. (a) Do you think that all proteins would be precipitated out to the same extent by the same concentration of the same electrolyte? (b) If a protein has been salted out, are the protein-protein interactions stronger or weaker than they were before the electrolyte was added? (c) A friend of yours who is taking a biochemistry class says that salting out works because the waters of hydration that surround the protein prefer to surround the electrolyte as the electrolyte is added; therefore, the protein's hydration shell is stripped away, leading to protein precipitation. Another friend of yours in the same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation. Discuss these two hypotheses. What kind of measurements would you need to make to distinguish between these two hypotheses?

Indicate the type of solute-solvent interaction (Section 11.2\()\) that should be most important in each of the following solutions: (a) \(\mathrm{CCl}_{4}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),\) , ( b ) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water, (c) \(\mathrm{KBr}\) in water, \((\mathbf{d}) \mathrm{HCl}\) in acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\)
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