Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 46

Assuming that air contains 78 percent \(\mathrm{N}_{2}, 21\) percent \(\mathrm{O}_{2},\) and 1 percent \(\mathrm{Ar},\) all by volume, how many \(\mathrm{mol}-\) ecules of each type of gas are present in \(1.0 \mathrm{~L}\) of air at STP?

Short Answer

Expert verified
The number of molecules of each type of gas present in 1.0 L of air at STP are \(2.09 \times 10^{22}\) molecules of Nitrogen, \(5.64 \times 10^{21}\) molecules of Oxygen and \(2.68 \times 10^{20}\) molecules of Argon.

Step by step solution

01

Determine the Amount of Each Gas

First, find out the volume of each of the three gases in air, this is achieved by multiplying each percentage given by the total volume of the air, in this case, 1.0 L. Therefore, For Nitrogen (\(N_2\)), the volume is \(0.78 \times 1.0 L = 0.78 L\),For Oxygen (\(O_2\)), the volume is \(0.21 \times 1.0 L = 0.21 L\), andFor Argon (\(Ar\)), the volume is \(0.01 \times 1.0 L = 0.01 L\).
02

Calculate the Moles of Each Gas

To find out the moles of gas, the formula derived from Avogadro’s law is utilized, which is volumegas / 22.4 (the volume per mole at STP). Hence,For \(N_2\), the moles are \(0.78 L / 22.4 L/mol = 0.0348 moles\),For \(O_2\), the moles are \(0.21 L / 22.4 L/mol = 0.00938 moles\), andFor \(Ar\), the moles are \(0.01 L / 22.4 L/mol = 0.000446 moles\).
03

Calculate the Number of Molecules

To convert moles to molecules, you multiply by Avogadro’s number (\(6.022 \times 10^{23}\) molecules/mol). Therefore,For \(N_2\), the number of molecules will be \(0.0348 moles \times 6.022 \times 10^{23} molecules/mol = 2.09 \times 10^{22}\) molecules,For \(O_2\), the number of molecules is \(0.00938 moles \times 6.022 \times 10^{23} molecules/mol = 5.64 \times 10^{21}\) molecules, andFor \(Ar\), the number of molecules is \(0.000446 moles \times 6.022 \times 10^{23} molecules/mol = 2.68 \times 10^{20}\) molecules.

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

The volume of a sample of pure HCl gas was \(189 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\) and \(108 \mathrm{mmHg}\). It was completely dissolved in about \(60 \mathrm{~mL}\) of water and titrated with an \(\mathrm{NaOH}\) solution; \(15.7 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution were required to neutralize the \(\mathrm{HCl}\). Calculate the molarity of the \(\mathrm{NaOH}\) solution.

Why do astronauts have to wear protective suits when they are on the surface of the moon?

Nickel forms a gaseous compound of the formula \(\mathrm{Ni}(\mathrm{CO})_{x}\). What is the value of \(x\) given the fact that under the same conditions of temperature and pressure methane \(\left(\mathrm{CH}_{4}\right)\) effuses 3.3 times faster than the compound?

Would it be easier to drink water with a straw on top of Mt. Everest or at the foot? Explain.

Air entering the lungs ends up in tiny sacs called alveoli. It is from the alveoli that oxygen diffuses into the blood. The average radius of the alveoli is \(0.0050 \mathrm{~cm}\) and the air inside contains 14 percent oxygen. Assuming that the pressure in the alveoli is 1.0 atm and the temperature is \(37^{\circ} \mathrm{C},\) calculate the number of oxygen molecules in one of the alveoli.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

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