World of Chemistry
World of Chemistry
7th Edition
ISBN: 9780618562763
Author: Steven S. Zumdahl
Publisher: Houghton Mifflin College Div
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Chapter 15, Problem 29A

(a)

Interpretation Introduction

Interpretation:

The number of moles in 10.2 mL of 0.451 M AlCl3 solution needs to be determined.

Concept Introduction:

The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  M=molarity=numberofmolesofsolutevolumeofthesolutionM=nV(L)

Where, n is the number of moles,

L is the volume of the solution.

(a)

Expert Solution
Check Mark

Answer to Problem 29A

  4.6×103mol

Explanation of Solution

Given information:

The volume is, V=10.2 mL=10.2×103L

The molarity is, M=0.451M

Calculation:

The molarity formula is,

  M=molarity=numberofmolesofsolutevolumeofthesolution

By substituting the given values in the formula and get the number of moles of ions,

  n=molarity×volume=0.451molL×10.2×10-3L=4.6×10-3moles

The number of moles of solute is 4.6×103moles .

(b)

Interpretation Introduction

Interpretation:

The number of moles of 5.51 L of 0.103 M Na3PO4solution needs to be determined.

Concept Introduction:

The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  M=molarity=numberofmolesofsolutevolumeofthesolutionM=nV(L)

Where, n is the number of moles,

L is the volume of the solution.

(b)

Expert Solution
Check Mark

Answer to Problem 29A

  0.567 mol

Explanation of Solution

Given information:

The volume is, V=0.103L

The molarity is, M=5.51M

Formula used:

The molarity formula is,

  M=molarity=numberofmolesofsolutevolumeofthesolution

Solution:

By substituting the given values in the formula and get the number of moles of ions,

  M=molarity=numberofmolesofsolutevolumeofthesolutionn=molarity×volume=5.51molL×0.103L=0.567moles

The number of moles of solute is 0.567 moles .

(c)

Interpretation Introduction

Interpretation:

The number of moles of 1.75 mL of 1.25 M CuCl2 solution needs to be determined.

Concept Introduction:

The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  M=molarity=numberofmolesofsolutevolumeofthesolutionM=nV(L)

Where, n is the number of moles,

L is the volume of the solution.

(c)

Expert Solution
Check Mark

Answer to Problem 29A

  2.18×103moles

Explanation of Solution

Given information:

The volume is, V=1.75mL=1.75×10-3L

The molarity is, M=1.25M

Calculation:

The molarity formula is,

  M=molarity=numberofmolesofsolutevolumeofthesolution

By substituting the given values in the formula and get the number of moles of ions,

  M=molarity=numberofmolesofsolutevolumeofthesolutionn=molarity×volume=1.25MmolL×1.75×10-3L=2.18×10-3moles

The number of moles of solute is 2.18×103moles .

(d)

Interpretation Introduction

Interpretation:

The number of moles of 25.2 mL of 0.00157 M Ca(OH)2 solution needs to be determined.

Concept Introduction:

The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  M=molarity=numberofmolesofsolutevolumeofthesolutionM=nV(L)

Where, n is the number of moles,

L is the volume of the solution.

(d)

Expert Solution
Check Mark

Answer to Problem 29A

  3.956×105moles

Explanation of Solution

Given information:

The volume is, V=25.20 mL=25.20×103L

The molarity is, M=0.00157 M

Calculation:

The molarity formula is,

  M=molarity=numberofmolesofsolutevolumeofthesolution

By substituting the given values in the formula and get the number of moles of ions,

  M=molarity=numberofmolesofsolutevolumeofthesolutionn=molarity×volume=0.00157molL×25.20×10-3L=3.956×10-5moles

The number of moles of solute is 3.956×105moles .

Chapter 15 Solutions

World of Chemistry

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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