World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
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Chapter 6, Problem 26A

(a)

Interpretation Introduction

Interpretation:

The number of molecules present in the samples needs to be predicted.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

(a)

Expert Solution
Check Mark

Answer to Problem 26A

Number of molecules of CO in 6.37 mol = 3.836 ×1022 molecules

Explanation of Solution

Number of molecules in 6.37 mol of carbon monoxide

Number of carbon monoxide molecules = Number of moles × Avogadro number

                                 = 6.37 mol × (6.022140857 × 1023 molecules / mol)

                                    = 38.361 × 1023 molecules

                                    = 3.836 ×1022 molecules

(b)

Interpretation Introduction

Interpretation:

The number of molecules present in the samples needs to be predicted.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

(b)

Expert Solution
Check Mark

Answer to Problem 26A

Number of molecules of CO in 6.37 g = 1.37 × 1023 molecules

Explanation of Solution

Number of molecules in 6.37 g carbon monoxide

 Molecular formula = CO            

Molecular mass = 28.01 g/ mol  

Given,

Mass of CO = 6.37g

The calculation of moles is shown below:

  Number of moles of CO = Mass of CO Molecular mass of CO                                     = 6.37 g28.01 g/ mol                                     = 0.227418779 moles

Number of CO molecules = Number of moles × Avogadro number

                                    = 0.227418779 mol × (6.022140857 × 1023 molecules / mol)

                                    = 1.37 ×1023 molecules

(c)

Interpretation Introduction

Interpretation:

The number of molecules present in the samples needs to be predicted.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

(c)

Expert Solution
Check Mark

Answer to Problem 26A

Number of molecules of H2O in 2.62 ×10-6 g = 8.76× 1016 molecules

Explanation of Solution

Number of molecules in 2.62×10-6 g water

Molecular formula = H2O           

 Molecular mass = 18g/ mol  

Given,

Mass of H2O = 2.62×10-6 g

The calculation of moles is shown below:

  Number of moles of H2O = Mass of H2Molecular mass of H2O                                     = 2.62×10-6 g18 g/ mol                                     = 1.4555×10-7 moles

Number of H2O molecules = Number of moles × Avogadro number

                                    = 1.4555 ×10-7 mol × (6.022140857 × 1023 molecules / mol)

                                    = 8.76 ×1016 molecules

(d)

Interpretation Introduction

Interpretation:

The number of molecules present in the samples needs to be predicted.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

(d)

Expert Solution
Check Mark

Answer to Problem 26A

Number of molecules of H2O in 2.62× 10-6 mol = 1.58 ×1018 molecules

Explanation of Solution

 Number of molecules in 2.62 × 10-6 mol of water

Number of water molecules = Number of moles × Avogadro number

                                    = 2.62 × 10-6 mol × (6.022140857 × 1023 molecules / mol)

                                    = 15.78 × 1017 molecules

                                    = 1.58 ×1018 molecules

(e)

Interpretation Introduction

Interpretation:

The number of molecules present in the samples needs to be predicted.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

(e)

Expert Solution
Check Mark

Answer to Problem 26A

Number of molecules of C6H6 in 5.23 g = 4.029× 1022 molecules

Explanation of Solution

Number of molecules in 5.23 g benzene

Molecular formula = C6H6            

Molecular mass = 78.11 g/ mol  

Given,

Mass of C6H6= 5.23 g

The calculation of moles is shown below:

  Number of moles of C6H6 = Mass of C6H6  Molecular mass of C6H6                                      = 5.23 g78.11 g/ mol                                     =0.0669 moles

Number of C6H6 molecules = Number of moles × Avogadro number

                                    = 0.0669 mol × (6.022140857 × 1023 molecules / mol)

                                    = 4.029 ×1022 molecules

Chapter 6 Solutions

World of Chemistry, 3rd edition

Ch. 6.2 - Prob. 5RQCh. 6.2 - Prob. 6RQCh. 6.3 - Prob. 1RQCh. 6.3 - Prob. 2RQCh. 6.3 - Prob. 3RQCh. 6.3 - Prob. 4RQCh. 6.3 - Prob. 5RQCh. 6.3 - Prob. 6RQCh. 6 - Prob. 1ACh. 6 - Prob. 2ACh. 6 - Prob. 3ACh. 6 - Prob. 4ACh. 6 - Prob. 5ACh. 6 - Prob. 6ACh. 6 - Prob. 7ACh. 6 - Prob. 8ACh. 6 - Prob. 9ACh. 6 - Prob. 10ACh. 6 - Prob. 11ACh. 6 - Prob. 12ACh. 6 - Prob. 13ACh. 6 - Prob. 14ACh. 6 - Prob. 15ACh. 6 - Prob. 16ACh. 6 - Prob. 17ACh. 6 - Prob. 18ACh. 6 - Prob. 19ACh. 6 - Prob. 20ACh. 6 - Prob. 21ACh. 6 - Prob. 22ACh. 6 - Prob. 23ACh. 6 - Prob. 24ACh. 6 - Prob. 25ACh. 6 - Prob. 26ACh. 6 - Prob. 27ACh. 6 - Prob. 28ACh. 6 - Prob. 29ACh. 6 - Prob. 30ACh. 6 - Prob. 31ACh. 6 - Prob. 32ACh. 6 - Prob. 33ACh. 6 - Prob. 34ACh. 6 - Prob. 35ACh. 6 - Prob. 36ACh. 6 - Prob. 37ACh. 6 - Prob. 38ACh. 6 - Prob. 39ACh. 6 - Prob. 40ACh. 6 - Prob. 41ACh. 6 - Prob. 42ACh. 6 - Prob. 43ACh. 6 - Prob. 44ACh. 6 - Prob. 45ACh. 6 - Prob. 46ACh. 6 - Prob. 47ACh. 6 - Prob. 48ACh. 6 - Prob. 49ACh. 6 - Prob. 50ACh. 6 - Prob. 51ACh. 6 - Prob. 52ACh. 6 - Prob. 53ACh. 6 - Prob. 54ACh. 6 - Prob. 55ACh. 6 - Prob. 56ACh. 6 - Prob. 57ACh. 6 - Prob. 58ACh. 6 - Prob. 59ACh. 6 - Prob. 60ACh. 6 - Prob. 61ACh. 6 - Prob. 62ACh. 6 - Prob. 63ACh. 6 - Prob. 64ACh. 6 - Prob. 65ACh. 6 - Prob. 66ACh. 6 - Prob. 67ACh. 6 - Prob. 68ACh. 6 - Prob. 69ACh. 6 - Prob. 70ACh. 6 - Prob. 1STPCh. 6 - Prob. 2STPCh. 6 - Prob. 3STPCh. 6 - Prob. 4STPCh. 6 - Prob. 5STPCh. 6 - Prob. 6STPCh. 6 - Prob. 7STPCh. 6 - Prob. 8STPCh. 6 - Prob. 9STPCh. 6 - Prob. 10STPCh. 6 - Prob. 11STPCh. 6 - Prob. 12STP
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Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY