General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
Question
Book Icon
Chapter 4, Problem 4.60QP

(a)

Interpretation Introduction

Interpretation:

Mass of solute needed to make 2.50×102mL of 0.100M of solution has to be identified.

Concept introduction:

 Molarity of the solution=MolesofsoluteVolumeinliter

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 Mole=Concentration(M)×volume(L)

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 Mass=Mole×Molarmass

(a)

Expert Solution
Check Mark

Explanation of Solution

Given,

    The concentration of solution is 0.100M

    The volume of solution is 2.50×102mL

The number of mole of cesium iodide can be calculated by using

Mole=Concentration(M)×volume(L)

Number of Moles=0.100M×2.50×102×103L=0.025mol

The mass of cesium iodide can be calculated by using,

From the standard data molar mass of CsI=259.8g/mol

=0.025mol×259.8g/mol=6.495g=6.50g

The 6.50g of cesium iodide is used to  2.50×102mL of a 0.100M of solution.

(b)

Interpretation Introduction

Interpretation:

Mass of solute needed to make 2.50×102mL of 0.100M of solution has to be identified.

Concept introduction:

 Molarity of the solution=MolesofsoluteVolumeinliter

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 Mole=Concentration(M)×volume(L)

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 Mass=Mole×Molarmass

(b)

Expert Solution
Check Mark

Explanation of Solution

Given,

    The concentration of solution is 0.100M

    The volume of solution is 2.50×102mL

The number of mole of H2SO4 can be calculated by using

Mole=Concentration(M)×volume(L)

Number of Moles=0.100M×2.50×102×103L=0.025mol

The mass of H2SO4 can be calculated by using,

From the standard data molar mass of H2SO4=98.086g/mol

=0.025mol×98.086g/mol=2.45g

The 2.45g of H2SO4 is used to  2.50×102mL of a 0.100M of solution.

(c)

Interpretation Introduction

Interpretation:

Mass of solute needed to make 2.50×102mL of a 0.100M of solution has to be identified.

Concept introduction:

 Molarity of the solution=MolesofsoluteVolumeinliter

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 Mole=Concentration(M)×volume(L)

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 Mass=Mole×Molarmass

(c)

Expert Solution
Check Mark

Explanation of Solution

Given,

    The concentration of solution is 0.100M

    The volume of solution is 2.50×102mL

The number of mole can be calculated by using

Mole=Concentration(M)×volume(L)

Number of Moles=0.100M×2.50×102×103L=0.025mol

The mass of Na2CO3 can be calculated by using,

From the standard data molar mass of Na2CO3=105.99g/mol

=0.025mol×105.99g/mol=2.649g

The 2.649g of Na2CO3 is used to  2.50×102mL of a 0.100M of solution.

(d)

Interpretation Introduction

Interpretation:

Mass of solute needed to make 2.50×102mL of a 0.100M of solution has to be identified.

Concept introduction:

 Molarity of the solution=MolesofsoluteVolumeinliter

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 Mole=Concentration(M)×volume(L)

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 Mass=Mole×Molarmass

(d)

Expert Solution
Check Mark

Explanation of Solution

Given,

    The concentration of solution is 0.100M

    The volume of solution is 2.50×102mL

The number of mole can be calculated by using

Mole=Concentration(M)×volume(L)

Number of Moles=0.100M×2.50×102×103L=0.025mol

The mass of K2Cr2O7 can be calculated by using,

From the standard data molar mass of K2Cr2O7=294.2g/mol

=0.025mol×294.2g/mol=7.355g

The 7.355g of K2Cr2O7 is used to  2.50×102mL of a 0.100M of solution.

(e)

Interpretation Introduction

Interpretation:

Mass of solute needed to make 2.50×102mL of a 0.100M of solution has to be identified.

Concept introduction:

 Molarity of the solution=MolesofsoluteVolumeinliter

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 Mole=Concentration(M)×volume(L)

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 Mass=Mole×Molarmass

(e)

Expert Solution
Check Mark

Explanation of Solution

Given,

    The concentration of solution is 0.100M

    The volume of solution is 2.50×102mL

The number of mole can be calculated by using

Mole=Concentration(M)×volume(L)

Number of Moles=0.100M×2.50×102×103L=0.025mol

The mass of KMnO4 can be calculated by using,

From the standard data molar mass of KMnO4=158.04g/mol

=0.025mol×158.04g/mol=3.95g

The 3.95g of KMnO4 is used to  2.50×102mL of a 0.100M of solution.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 4 Solutions

General Chemistry

Ch. 4.5 - Prob. 1PECh. 4.5 - Prob. 2PECh. 4.5 - Prob. 3PECh. 4.5 - Prob. 1RCCh. 4.6 - Prob. 1PECh. 4.6 - Prob. 1RCCh. 4.6 - Prob. 2PECh. 4.6 - Prob. 3PECh. 4 - Prob. 4.1QPCh. 4 - Prob. 4.2QPCh. 4 - Prob. 4.3QPCh. 4 - 4.4 What is the difference between the following...Ch. 4 - 4.5 Water is an extremely weak electrolyte and...Ch. 4 - Prob. 4.6QPCh. 4 - Prob. 4.7QPCh. 4 - 4.8 Which of the following diagrams best...Ch. 4 - Prob. 4.9QPCh. 4 - Prob. 4.10QPCh. 4 - Prob. 4.11QPCh. 4 - Prob. 4.12QPCh. 4 - Prob. 4.13QPCh. 4 - Prob. 4.14QPCh. 4 - Prob. 4.15QPCh. 4 - Prob. 4.16QPCh. 4 - Prob. 4.17QPCh. 4 - Prob. 4.18QPCh. 4 - Prob. 4.19QPCh. 4 - Prob. 4.20QPCh. 4 - 4.21 Write ionic and net ionic equations for the...Ch. 4 - Prob. 4.22QPCh. 4 - Prob. 4.23QPCh. 4 - Prob. 4.24QPCh. 4 - Prob. 4.25QPCh. 4 - Prob. 4.26QPCh. 4 - Prob. 4.27QPCh. 4 - Prob. 4.28QPCh. 4 - Prob. 4.29QPCh. 4 - Prob. 4.30QPCh. 4 - Prob. 4.31QPCh. 4 - Prob. 4.32QPCh. 4 - Prob. 4.33QPCh. 4 - Prob. 4.34QPCh. 4 - Prob. 4.35QPCh. 4 - Prob. 4.36QPCh. 4 - Prob. 4.37QPCh. 4 - Prob. 4.38QPCh. 4 - 4.39 For the complete redox reactions given here,...Ch. 4 - Prob. 4.40QPCh. 4 - Prob. 4.41QPCh. 4 - Prob. 4.42QPCh. 4 - Prob. 4.43QPCh. 4 - Prob. 4.44QPCh. 4 - Prob. 4.45QPCh. 4 - Prob. 4.46QPCh. 4 - Prob. 4.47QPCh. 4 - Prob. 4.48QPCh. 4 - Prob. 4.49QPCh. 4 - Prob. 4.50QPCh. 4 - Prob. 4.51QPCh. 4 - Prob. 4.52QPCh. 4 - Prob. 4.53QPCh. 4 - Prob. 4.54QPCh. 4 - Prob. 4.55QPCh. 4 - Prob. 4.56QPCh. 4 - Prob. 4.57QPCh. 4 - Prob. 4.58QPCh. 4 - Prob. 4.59QPCh. 4 - Prob. 4.60QPCh. 4 - Prob. 4.61QPCh. 4 - Prob. 4.62QPCh. 4 - Prob. 4.63QPCh. 4 - Prob. 4.64QPCh. 4 - Prob. 4.65QPCh. 4 - Prob. 4.66QPCh. 4 - Prob. 4.67QPCh. 4 - Prob. 4.68QPCh. 4 - Prob. 4.69QPCh. 4 - 4.70 Distilled water must be used in the...Ch. 4 - 4.71 If 30.0 mL of 0.150 M CaCl2 is added to 15.0...Ch. 4 - Prob. 4.72QPCh. 4 - Prob. 4.73QPCh. 4 - Prob. 4.74QPCh. 4 - Prob. 4.75QPCh. 4 - Prob. 4.76QPCh. 4 - Prob. 4.77QPCh. 4 - Prob. 4.78QPCh. 4 - Prob. 4.79QPCh. 4 - Prob. 4.80QPCh. 4 - Prob. 4.81QPCh. 4 - Prob. 4.82QPCh. 4 - Prob. 4.83QPCh. 4 - Prob. 4.84QPCh. 4 - Prob. 4.85QPCh. 4 - Prob. 4.86QPCh. 4 - Prob. 4.87QPCh. 4 - Prob. 4.88QPCh. 4 - Prob. 4.89QPCh. 4 - Prob. 4.90QPCh. 4 - Prob. 4.91QPCh. 4 - Prob. 4.92QPCh. 4 - Prob. 4.93QPCh. 4 - 4.74 The molecular formula of malonic acid is...Ch. 4 - Prob. 4.95QPCh. 4 - Prob. 4.96QPCh. 4 - Prob. 4.97QPCh. 4 - Prob. 4.98QPCh. 4 - Prob. 4.99QPCh. 4 - Prob. 4.100QPCh. 4 - Prob. 4.101QPCh. 4 - Prob. 4.102QPCh. 4 - 4.103 These are common household compounds: table...Ch. 4 - Prob. 4.104QPCh. 4 - Prob. 4.105QPCh. 4 - Prob. 4.106QPCh. 4 - 4.107 A number of metals are involved in redox...Ch. 4 - Prob. 4.108QPCh. 4 - Prob. 4.109QPCh. 4 - Prob. 4.110QPCh. 4 - Prob. 4.111QPCh. 4 - Prob. 4.112QPCh. 4 - Prob. 4.114SPCh. 4 - Prob. 4.115SPCh. 4 - Prob. 4.116SPCh. 4 - Prob. 4.117SPCh. 4 - Prob. 4.118SP
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