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
Question
Book Icon
Chapter 6.1, Problem 4RQ
Interpretation Introduction

Interpretation: The given table needs to be completed for mass of samples, number of moles and atoms in sample.

Concept Introduction: In 1 mol of a substance there are 6.022 × 1023 particles. This is known as Avogadro’s number and represented by symbol NA.

Expert Solution & Answer
Check Mark

Answer to Problem 4RQ

  Number of moles of Cu=0.674 mol

  Number of atoms or molecules of Cu=4.05×1023 atoms

  mass of K=33.38 g

  Number of atoms or molecules of K=5.15×1023

  mass of Ar=23.7g

  Number of atoms or molecules of Ar=1.185×102

  Number of moles of Cs=0.0144×10-3

  Number of atoms or molecules of Cs=0.086×1020

  mass of He=19.04 g

  Number of atoms or molecules of He=28.66×1023

  Mass of Si=0.179g

  Number of atoms of Si=0.0639 mol

Explanation of Solution

The number of moles can be calculated from mass and molar mass as follows:

  Number of moles=Given massAtomic mass

Putting the values,

  Number of moles of Cu=42.8 g63.5 g/mol=0.674 mol

The number of atoms or molecules in the given mass of an atom can be calculated as follows:

  Number of molecules or atoms of Cu=moles×Avogadro's number

  =0.674×6.022×1023 atoms

  =4.05×1023 atoms

The mass of K can be calculated as follows:

  Number of moles of K=Given massAtomic mass

  0.856=xgm39

  x=0.856×39

  x=33.38gm

Now, number of atoms of K can be calculated as follows:

  Number of molecules or atoms of K=moles×Avogadro's number

  =0.856×6.022×1023

  =5.15×1023

The number of moles of Ar can be calculated as follows:

  Moles=number of atoms or molecules of ArAvogadro's number

  =7.14×10256.022×1023

  =1.185×102

Now, from number of moles, mass of Ar can be calculated as follows:

  Number of moles of Ar=Given massAtomic mass

  1.185×102=xgm20

  x=1.185×102×20

  x=23.7gm

The number of moles can be calculated as follows:

  Number of moles=Given massAtomic mass

  =1.92×108132.9

  =0.0144×103

Now, number of molecules can be calculated as follows:

  Number of molecules or atoms=moles×Avogadro's number

  =0.0144×103×6.022×1023

  =0.086×1020

The mass of He can be calculated as follows:

  Number of moles=Given massAtomic mass

  4.76=xgm4

  x=4.76×4

  x=19.04gm

The number of molecules can be calculated as follows:

  Number of molecules or atoms=moles×Avogadro's number

  =4.76×103×6.022×1023

  =28.66×1023

The number of moles of Si can be calculated as follows:

  nSi=3.85×10226.022×1023

  =0.639×101

  =0.0639

Now, mass can be calculated as follows:

  Number of moles=Given massAtomic mass

  0.0639=xgm28.08

  x=0.0639×28.08

  x=1.79gm

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
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY