EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
Question
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Chapter 14, Problem 22PE

(a)

Interpretation Introduction

Interpretation:

Grams of solute in 1.20 L of 18 M H2SO4 have to be determined.

Concept Introduction:

Molarity is amount of solute per 1 L of solution. Formula for molarity is as follows:

  Molarity of solution=Moles of soluteVolume (L) of solution

(a)

Expert Solution
Check Mark

Answer to Problem 22PE

2118.5 g is present in 1.20 L of 18 M H2SO4.

Explanation of Solution

Formula for molarity of H2SO4 solution is as follows:

  Molarity of H2SO4 solution=Moles of H2SO4Volume (L) of H2SO4 solution        (1)

Rearrange equation (1) for moles of H2SO4.

  Moles of H2SO4=[(Molarity of H2SO4 solution)(Volume (L) of H2SO4 solution)]        (2)

Substitute 18 M for molarity and 1.20 L for volume of H2SO4 solution in equation (2).

  Moles of H2SO4=(18 M)(1.20 L)=21.6 mol

Formula for mass of H2SO4 is as follows:

  Mass of H2SO4=(Moles of H2SO4)(Molar mass of H2SO4)        (3)

Substitute 21.6 mol for moles and 98.079 g/mol for molar mass of H2SO4 in equation (3).

  Mass of H2SO4=(21.6 mol)(98.079 g/mol)=2118.5 g

Hence, 2118.5 g is present in 1.20 L of 18 M H2SO4.

(b)

Interpretation Introduction

Interpretation:

Grams of solute in 27.5 mL of 1.50 M KMnO4 have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 22PE

6.52 g is present in 27.5 mL of 1.50 M KMnO4.

Explanation of Solution

Formula for molarity of KMnO4 solution is as follows:

  Molarity of KMnO4 solution=Moles of KMnO4Volume (L) of KMnO4 solution        (4)

Rearrange equation (4) for moles of KMnO4.

  Moles of KMnO4=[(Molarity of KMnO4 solution)(Volume (L) of KMnO4 solution)]        (5)

Substitute 1.50 M for molarity and 27.5 mL for volume of KMnO4 solution in equation (5).

  Moles of KMnO4=(1.50 M)(27.5 mL)(103 L1 mL)=0.04125 mol

Formula for mass of KMnO4 is as follows:

  Mass of KMnO4=[(Moles of KMnO4)(Molar mass of KMnO4)]        (6)

Substitute 0.04125 mol for moles and 158.034 g/mol for molar mass of KMnO4 in equation (6).

  Mass of KMnO4=(0.04125 mol)(158.034 g/mol)=6.52 g

Hence, 6.52 g is present in 27.5 mL of 1.50 M KMnO4.

(c)

Interpretation Introduction

Interpretation:

Grams of solute in 120 mL of 0.025 M Fe2(SO4)3 have to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 22PE

1.20 g is present in 120 mL of 0.025 M Fe2(SO4)3.

Explanation of Solution

Formula for molarity of Fe2(SO4)3 solution is as follows:

  Molarity of Fe2(SO4)3 solution=Moles of Fe2(SO4)3Volume (L) of Fe2(SO4)3 solution        (7)

Rearrange equation (7) for moles of Fe2(SO4)3.

  Moles of Fe2(SO4)3=[(Molarity of Fe2(SO4)3 solution)(Volume (L) of Fe2(SO4)3 solution)]        (8)

Substitute 0.025 M for molarity and 120 mL for volume of Fe2(SO4)3 solution in equation (8).

  Moles of Fe2(SO4)3=(0.025 M)(120 mL)(103 L1 mL)=0.003 mol

Formula for mass of Fe2(SO4)3 is as follows:

  Mass of Fe2(SO4)3=[(Moles of Fe2(SO4)3)(Molar mass of Fe2(SO4)3)]        (9)

Substitute 0.003 mol for moles and 399.88 g/mol for molar mass of Fe2(SO4)3 in equation (9).

  Mass of Fe2(SO4)3=(0.003 mol)(399.88 g/mol)=1.20 g

Hence, 1.20 g is present in 120 mL of 0.025 M Fe2(SO4)3.

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Chapter 14 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 14.5 - Prob. 14.11PCh. 14.5 - Prob. 14.12PCh. 14 - Prob. 1RQCh. 14 - Prob. 2RQCh. 14 - Prob. 3RQCh. 14 - Prob. 4RQCh. 14 - Prob. 5RQCh. 14 - Prob. 6RQCh. 14 - Prob. 7RQCh. 14 - Prob. 8RQCh. 14 - Prob. 9RQCh. 14 - Prob. 10RQCh. 14 - Prob. 11RQCh. 14 - Prob. 12RQCh. 14 - Prob. 13RQCh. 14 - Prob. 14RQCh. 14 - Prob. 15RQCh. 14 - Prob. 16RQCh. 14 - Prob. 17RQCh. 14 - Prob. 18RQCh. 14 - Prob. 19RQCh. 14 - Prob. 20RQCh. 14 - Prob. 21RQCh. 14 - Prob. 22RQCh. 14 - Prob. 23RQCh. 14 - Prob. 24RQCh. 14 - Prob. 25RQCh. 14 - Prob. 26RQCh. 14 - Prob. 27RQCh. 14 - Prob. 28RQCh. 14 - Prob. 29RQCh. 14 - Prob. 30RQCh. 14 - Prob. 31RQCh. 14 - Prob. 32RQCh. 14 - Prob. 33RQCh. 14 - Prob. 34RQCh. 14 - Prob. 35RQCh. 14 - Prob. 37RQCh. 14 - Prob. 38RQCh. 14 - Prob. 39RQCh. 14 - Prob. 40RQCh. 14 - Prob. 41RQCh. 14 - Prob. 42RQCh. 14 - Prob. 1PECh. 14 - Prob. 2PECh. 14 - Prob. 3PECh. 14 - Prob. 4PECh. 14 - Prob. 5PECh. 14 - Prob. 6PECh. 14 - Prob. 7PECh. 14 - Prob. 8PECh. 14 - Prob. 9PECh. 14 - Prob. 10PECh. 14 - Prob. 11PECh. 14 - Prob. 12PECh. 14 - Prob. 13PECh. 14 - Prob. 14PECh. 14 - Prob. 15PECh. 14 - Prob. 16PECh. 14 - Prob. 17PECh. 14 - Prob. 18PECh. 14 - Prob. 19PECh. 14 - Prob. 20PECh. 14 - Prob. 21PECh. 14 - Prob. 22PECh. 14 - Prob. 23PECh. 14 - Prob. 24PECh. 14 - Prob. 25PECh. 14 - Prob. 26PECh. 14 - Prob. 27PECh. 14 - Prob. 28PECh. 14 - Prob. 29PECh. 14 - Prob. 30PECh. 14 - Prob. 31PECh. 14 - Prob. 32PECh. 14 - Prob. 33PECh. 14 - Prob. 34PECh. 14 - Prob. 35PECh. 14 - Prob. 36PECh. 14 - Prob. 37PECh. 14 - Prob. 38PECh. 14 - Prob. 39PECh. 14 - Prob. 40PECh. 14 - Prob. 41PECh. 14 - Prob. 42PECh. 14 - Prob. 44PECh. 14 - Prob. 45PECh. 14 - Prob. 46PECh. 14 - Prob. 47PECh. 14 - Prob. 48PECh. 14 - Prob. 49PECh. 14 - Prob. 50PECh. 14 - Prob. 51PECh. 14 - Prob. 52PECh. 14 - Prob. 53AECh. 14 - Prob. 54AECh. 14 - Prob. 55AECh. 14 - Prob. 56AECh. 14 - Prob. 57AECh. 14 - Prob. 58AECh. 14 - Prob. 59AECh. 14 - Prob. 60AECh. 14 - Prob. 61AECh. 14 - Prob. 62AECh. 14 - Prob. 63AECh. 14 - Prob. 65AECh. 14 - Prob. 66AECh. 14 - Prob. 67AECh. 14 - Prob. 68AECh. 14 - Prob. 69AECh. 14 - Prob. 70AECh. 14 - Prob. 71AECh. 14 - Prob. 72AECh. 14 - Prob. 73AECh. 14 - Prob. 74AECh. 14 - Prob. 75AECh. 14 - Prob. 76AECh. 14 - Prob. 77AECh. 14 - Prob. 78AECh. 14 - Prob. 79AECh. 14 - Prob. 80AECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Prob. 83AECh. 14 - Prob. 84AECh. 14 - Prob. 85AECh. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - Prob. 88AECh. 14 - Prob. 90AECh. 14 - Prob. 91AECh. 14 - Prob. 92AECh. 14 - Prob. 93AECh. 14 - Prob. 94AECh. 14 - Prob. 95AECh. 14 - Prob. 96AECh. 14 - Prob. 97AECh. 14 - Prob. 98AECh. 14 - Prob. 99CECh. 14 - Prob. 100CECh. 14 - Prob. 102CECh. 14 - Prob. 103CECh. 14 - Prob. 104CECh. 14 - Prob. 105CE
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