Log out Go to App
Go to App

Learning Materials

Features

Discover

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 101.3 kPa pressure and a body temperature of \(37^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) The mol percentage of CO2 in the air is ≈ \(4.142\%\). (b) The molarity of CO2 in the air is ≈ \(0.1054\,\text{M}\).

Step by step solution

01

(Step 1: Calculate the mole fraction of CO2)

First, we will need to convert the given volume percentage of CO2 to moles. As gases are mixed, they occupy separate volumes, and we can assume that 4.0% by volume means that 4.0% of the moles of the total gas mixture are CO2. Given that air is mainly composed of Nitrogen (78.1%), Oxygen(20.9%), Argon (0.93%), and CO2 (0.04%), we can normalize these percentages and obtain the mol percentages for these gases. For CO2: Mol percentage of CO2 = \(\frac{4.0\%}{78.1\% + 20.9\% + 0.93\% + 4.0 \%}\)
02

(Step 2: Calculate the mole fraction of CO2)

Calculate the mol percentage by dividing the given volume percentage by the sum of the volume percentages: Mol percentage of CO2 = \(\frac{4.0\%}{(78.1\% + 20.9\% + 0.93\% + 4.0 \%)} * 100\%\) Mol percentage of CO2 ≈ \(4.142\%\) This is the answer for part (a).
03

(Step 3: Calculate the moles of CO2 using the ideal gas equation)

Next, we will find the molarity of CO2, assuming 101.3 kPa pressure and a body temperature of \(37^\circ C\). We will use the ideal gas equation: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the ideal gas equation, we get: \(n = \frac{PV}{RT}\) To calculate the moles of CO2 in one liter of air, we have: P = 101.3 kPa (we are only interested in the CO2 portion, so we use the partial pressure) V = 1 L (one liter of air) R = 8.314 J/(mol*K) (ideal gas constant) T = 310.15 K (convert 37ºC to Kelvin by adding 273.15)
04

(Step 4: Calculate the moles of CO2 in 1L air)

We can plug these values into the equation in step 3: \(n_{CO2} = \frac{(4.142\% * 101.3\,\text{kPa}) * 1\,\text{L}}{8.314\,\text{J/(mol*K)} * 310.15\,\text{K}}\) \(n_{CO2} \approx 0.1054\,\text{mol}\)
05

(Step 5: Calculate the molarity of CO2)

Now, we can calculate the molarity (moles of solute per liter of solution) of CO2 in the air: Molarity = \(\frac{0.1054\,\text{mol}}{1\,\text{L}}\) Molarity ≈ \(0.1054\,\text{M}\) This is the answer for part (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q)\) (b) \(50.0 \mathrm{mg}\) of an aqueous solution that is $1.50 \mathrm{~m} \mathrm{NaCl}$, (c) \(75.0 \mathrm{~g}\) of an aqueous solution that is \(1.50 \%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.

A solution is made containing \(50.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(1000 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Calculate \((\mathbf{a})\) the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) (b) the mass percent of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), (c) the molality of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\).

If you compare the solubilities of the noble gases in water, you find that solubility increases from smallest atomic weight to largest, \(\mathrm{Ar}<\mathrm{Kr}<\mathrm{Xe}\). Which of the following statements is the best explanation? [Section 13.3] (a) The heavier the gas, the more it sinks to the bottom of the water and leaves room for more gas molecules at the top of the water. (b) The heavier the gas, the more dispersion forces it has, and therefore the more attractive interactions it has with water molecules. (c) The heavier the gas, the more likely it is to hydrogenbond with water. (d) The heavier the gas, the more likely it is to make a saturated solution in water.

Commercial concentrated aqueous ammonia is \(28 \% \mathrm{NH}_{3}\) by mass and has a density of \(0.90 \mathrm{~g} / \mathrm{mL}\). What is the molarity of this solution?

What is the osmotic pressure formed by dissolving \(50.0 \mathrm{mg}\) of acetylsalicylic acid $\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\( in \)0.100 \mathrm{~L}\( of water at \)37^{\circ} \mathrm{C} ?$
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free