General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
Question
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Chapter 4, Problem 4.108QP

(a)

Interpretation Introduction

Interpretation:

The required mass of KMnO4 to prepare a final solution should be calculated and reason for why not the very dilute solution are directly prepared should be explained.

Concept introduction:

Volumetric principle:

  • In the neutralization reaction, the volume and concentration of initial components are equal to the volume and concentration of the final components.
  • In the dilution of the chemical solutions are using volumetric principle.
  • In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

Mc×mLc=Md×mLdMc=InitialconcentrationmLc=InitialvolumeMd=dilutedconcentrationmLd=dilutedvolume

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

Molarity=No.molevolume(L)

Mole:

The mole of the solute is calculated by taken mass of solute divided by molar mass of the solute.

Mole=Mass(g)Molarmass(g)

(a)

Expert Solution
Check Mark

Answer to Problem 4.108QP

The required mass of KMnO4 to prepare a final solution is 3.29×10-5gKMnO4

Explanation of Solution

To record the given data,

Taken mass KMnO4 of in first dilution = 0.8214 g

Final Volume of KMnO4 in first dilution = 500 mL

Taken Volume of KMnO4 in second dilution = 2.000 mL

Final Volume of KMnO4 in second dilution = 1000 mL

Taken Volume of KMnO4 in third dilution= 10 mL

Final Volume of KMnO4 in third dilution = 250 mL

The taken mass and volumes of KMnO4 solutions are recorded as shown above.

Calculate the molarity of KMnO4 first diluted solution.

Molar mass of KMnO4 is 158.04 g

=0.8214gKMnO4×1molKMnO4158.04KMnO4=0.0051974mole

The molarity of KMnO4 is,

=0.00519740.500L=0.010395M

  • The taken mass of KMnO4 is divided by molar mass of KMnO4 to give the mole of taken KMnO4.
  • The calculated mole is dividing by volume of solution to give molarity of first diluted KMnO4 solution.
  • The molarity of first diluted KMnO4 solution is 0.010395M.

Calculate the molarity second diluted KMnO4 solution.

M1V1=M2V21000mL=0.0103952.000mL=0.0103952.000mL1000mL=2.079×10-5M

  • The calculated concentration and volume of first diluted KMnO4 solution are plugged in the above equation to give the concentration of second diluted KMnO4 solution.
  • The concentration of second diluted KMnO4 solution 2.079×10-5M

Calculate the molarity second third KMnO4 solution

M1V1=M2V2250mL=0.0000207910.0mL=0.00002079×10.0mL250mL=8.32×10-7M

  • The calculated concentration and volume of second diluted KMnO4 solution are plugged in the above equation to give the concentration of third diluted KMnO4 solution.
  • The concentration of third diluted KMnO4 solution 8.32×10-7M

Calculate the mass of KMnO4 in third solution

The mole of KMnO4 in third solution,

=8.32×10-7M1000mLsoln×250mL=2.08×10-7mole

The mass of KMnO4 in third solution is,

=2.08×10-7mole×158.04KMnO41molKMnO4=3.29×10-5gKMnO4

  • The calculated concentration and volume of third diluted KMnO4 solution are plugged in the above equation to give the mole of third diluted KMnO4 solution.
  • The mole of third diluted KMnO4 solution 2.08×10-7mole
  • The calculated mole of third diluted KMnO4 solution is multiplied by molar mass of KMnO4 to give mass of present in third diluted KMnO4 solution.
  • The mass of present in third diluted KMnO4 solution is 3.29×10-5gKMnO4

(b)

Interpretation Introduction

Interpretation:

The required mass of KMnO4 to prepare a final solution should be calculated and reason for why not the very dilute solution are directly prepared should be explained.

Concept introduction:

Volumetric principle:

  • In the neutralization reaction, the volume and concentration of initial components are equal to the volume and concentration of the final components.
  • In the dilution of the chemical solutions are using volumetric principle.
  • In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

Mc×mLc=Md×mLdMc=InitialconcentrationmLc=InitialvolumeMd=dilutedconcentrationmLd=dilutedvolume

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

Molarity=No.molevolume(L)

Mole:

The mole of the solute is calculated by taken mass of solute divided by molar mass of the solute.

Mole=Mass(g)Molarmass(g)

(b)

Expert Solution
Check Mark

Answer to Problem 4.108QP

The weight of solutes to prepare a very dilute solution is too small to directly weigh accurately.  So it results in more errors.  Therefore, the direct preparation of very dilute solution is not a good method.

Explanation of Solution

To record the given data,

Taken mass KMnO4 of in first dilution = 0.8214 g

Final Volume of KMnO4 in first dilution = 500 mL

Taken Volume of KMnO4 in second dilution = 2.000 mL

Final Volume of KMnO4 in second dilution = 1000 mL

Taken Volume of KMnO4 in third dilution= 10 mL

Final Volume of KMnO4 in third dilution = 250 mL

The taken mass and volumes of KMnO4 solutions are recorded as shown above.

Explain reason for why not the very dilute solution are directly prepared.

  • The weight of solutes to prepare a very dilute solution is too small to directly weigh accurately.  So, it results in more errors.  Therefore, the direct preparation of very dilute solution is not a good method.

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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
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