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 15, Problem 19A
Interpretation Introduction

Interpretation:

Number of moles of each ion per litre in aCaCl2 solution has to be calculated.

Concept Introduction:

Composition of a solution can be defined by expressing its concentration. The concentrations of solutions can be expressed in different ways, which are involved in the quantity of solute and the quantity of solution or solvent. Various methods are used to describe the concentration of the solution quantitatively. Some commonly used quantitative concentration terms are percent by mass, percent by volume, molarity, molality and mole fraction.

Molarity: Molarity is defined as the number of moles of solute present in one litre of the solution In expression, Molarity of the solution=Numberofmolesof the solute Volume of the solution(inlitre)

Again Numberofmolesofsolute=MassofthesoluteMolecularweightofthesolute

Expert Solution & Answer
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Answer to Problem 19A

0.221 moles of Ca2+ ions and 0.442 moles of Cl- ions are present per litre in a CaCl2 solution.

Explanation of Solution

  Molarity of the solution=Numberofmolesof the solute Volume of the solution(inlitre)

Data given: Molarity of CaCl2 solution = 0.221 M.

  Molarity of CaCl2 solution=Numberofmolesof CaCl2 Volume of CaCl2 solution(inlitre)Numberofmolesof CaCl2=Molarity of CaCl2solution×Volume of CaCl2 solution(inlitre)Numberofmolesof CaCl2=0.221mol/litre×1litre=0.221mole

  CaCl2Ca2++2Cl-

1 mole of CaCl2

gives 1 mole of World of Chemistry, 3rd edition, Chapter 15, Problem 19A , additional homework tip  1ion and 2 moles ofWorld of Chemistry, 3rd edition, Chapter 15, Problem 19A , additional homework tip  2ions. So, 0.221 mole of World of Chemistry, 3rd edition, Chapter 15, Problem 19A , additional homework tip  30.221 mole of World of Chemistry, 3rd edition, Chapter 15, Problem 19A , additional homework tip  4ions and 0.442 moles of World of Chemistry, 3rd edition, Chapter 15, Problem 19A , additional homework tip  5ions, 0.221 mole of World of Chemistry, 3rd edition, Chapter 15, Problem 19A , additional homework tip  6ions and 0.442 mole of World of Chemistry, 3rd edition, Chapter 15, Problem 19A , additional homework tip  7ions are present per litre in a CaCl2 solution.

Chapter 15 Solutions

World of Chemistry, 3rd edition

Ch. 15.2 - Prob. 5RQCh. 15.2 - Prob. 6RQCh. 15.2 - Prob. 7RQCh. 15.3 - Prob. 1RQCh. 15.3 - Prob. 2RQCh. 15.3 - Prob. 3RQCh. 15.3 - Prob. 4RQCh. 15.3 - Prob. 5RQCh. 15.3 - Prob. 6RQCh. 15.3 - Prob. 7RQCh. 15.3 - Prob. 8RQCh. 15 - Prob. 1ACh. 15 - Prob. 2ACh. 15 - Prob. 3ACh. 15 - Prob. 4ACh. 15 - Prob. 5ACh. 15 - Prob. 6ACh. 15 - Prob. 7ACh. 15 - Prob. 8ACh. 15 - Prob. 9ACh. 15 - Prob. 10ACh. 15 - Prob. 11ACh. 15 - Prob. 12ACh. 15 - Prob. 13ACh. 15 - Prob. 14ACh. 15 - Prob. 15ACh. 15 - Prob. 16ACh. 15 - Prob. 17ACh. 15 - Prob. 18ACh. 15 - Prob. 19ACh. 15 - Prob. 20ACh. 15 - Prob. 21ACh. 15 - Prob. 22ACh. 15 - Prob. 23ACh. 15 - Prob. 24ACh. 15 - Prob. 25ACh. 15 - Prob. 26ACh. 15 - Prob. 27ACh. 15 - Prob. 28ACh. 15 - Prob. 29ACh. 15 - Prob. 30ACh. 15 - Prob. 31ACh. 15 - Prob. 32ACh. 15 - Prob. 33ACh. 15 - Prob. 34ACh. 15 - Prob. 35ACh. 15 - Prob. 36ACh. 15 - Prob. 37ACh. 15 - Prob. 38ACh. 15 - Prob. 39ACh. 15 - Prob. 40ACh. 15 - Prob. 41ACh. 15 - Prob. 42ACh. 15 - Prob. 43ACh. 15 - Prob. 44ACh. 15 - Prob. 45ACh. 15 - Prob. 46ACh. 15 - Prob. 47ACh. 15 - Prob. 48ACh. 15 - Prob. 49ACh. 15 - Prob. 50ACh. 15 - Prob. 51ACh. 15 - Prob. 52ACh. 15 - Prob. 53ACh. 15 - Prob. 54ACh. 15 - Prob. 55ACh. 15 - Prob. 56ACh. 15 - Prob. 57ACh. 15 - Prob. 58ACh. 15 - Prob. 59ACh. 15 - Prob. 60ACh. 15 - Prob. 61ACh. 15 - Prob. 62ACh. 15 - Prob. 63ACh. 15 - Prob. 1STPCh. 15 - Prob. 2STPCh. 15 - Prob. 3STPCh. 15 - Prob. 4STPCh. 15 - Prob. 5STPCh. 15 - Prob. 6STPCh. 15 - Prob. 7STPCh. 15 - Prob. 8STPCh. 15 - Prob. 9STPCh. 15 - Prob. 10STPCh. 15 - Prob. 11STP
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY