EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
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Chapter 14, Problem 17PE

(a)

Interpretation Introduction

Interpretation:

Molarity of 0.25 mol of solute in 75.0 mL of solution has 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        (1)

(a)

Expert Solution
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Answer to Problem 17PE

Molarity of 0.25 mol of solute in 75.0 mL of solution is 3.33 M.

Explanation of Solution

Substitute 0.25 mol for moles of solute and 75.0 mL for volume of solution in equation (1).

  Molarity of solution=(0.25 mol75.0 mL)(1 mL103 L)=3.33 M

Hence, molarity of given solution is 3.33 M.

(b)

Interpretation Introduction

Interpretation:

Molarity of 1.75 mol of KBr in 0.75 L of solution has 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        (1)

(b)

Expert Solution
Check Mark

Answer to Problem 17PE

Molarity of 1.75 mol of KBr in 0.75 L of solution is 2.33 M.

Explanation of Solution

Formula for molarity of KBr solution is as follows:

  Molarity of KBr solution=Moles of KBrVolume (L) of KBr solution        (2)

Substitute 1.75 mol for moles of KBr and 0.75 L for volume of solution in equation (2).

  Molarity of KBr solution=1.75 mol0.75 L=2.33 M

Hence, molarity of given solution is 2.33 M.

(c)

Interpretation Introduction

Interpretation:

Molarity of 35.0 g of NaC2H3O2 in 1.25 L of solution has 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        (1)

(c)

Expert Solution
Check Mark

Answer to Problem 17PE

Molarity of 35.0 g of NaC2H3O2 in 1.25 L of solution is 0.3416 M.

Explanation of Solution

Formula for molarity of NaC2H3O2 solution is as follows:

  Molarity of NaC2H3O2 solution=Moles of NaC2H3O2Volume (L) of NaC2H3O2 solution        (3)

Formula for moles of NaC2H3O2 is as follows:

  Moles of NaC2H3O2=Mass of NaC2H3O2Molar mass of NaC2H3O2        (4)

Substitute 35.0 g for mass of NaC2H3O2 and 82.03 g/mol for molar mass of NaC2H3O2 in equation (4).

  Moles of NaC2H3O2=35.0 g82.03 g/mol=0.427 mol

Substitute 0.427 mol for moles and 1.25 L for volume of NaC2H3O2 solution in equation (3).

  Molarity of NaC2H3O2 solution=0.427 mol1.25 L=0.3416 M

Hence, molarity of given solution is 0.3416 M.

(d)

Interpretation Introduction

Interpretation:

Molarity of 75 g of CuSO45H2O in 1.0 L of solution has 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        (1)

(d)

Expert Solution
Check Mark

Answer to Problem 17PE

Molarity of 75 g of CuSO45H2O in 1.0 L of solution is 0.3004 M.

Explanation of Solution

Formula for molarity of CuSO45H2O solution is as follows:

  Molarity of CuSO45H2O solution=Moles of CuSO45H2OVolume (L) of CuSO45H2O solution        (5)

Formula for moles of CuSO45H2O is as follows:

  Moles of CuSO45H2O=Mass of CuSO45H2OMolar mass of CuSO45H2O        (6)

Substitute 75 g for mass of CuSO45H2O and 249.659 g/mol for molar mass of CuSO45H2O in equation (6).

  Moles of CuSO45H2O=75 g249.659 g/mol=0.3004 mol

Substitute 0.3004 mol for moles and 1.0 L for volume of CuSO45H2O solution in equation (3).

  Molarity of CuSO45H2O solution=0.3004 mol1.0 L=0.3004 M

Hence, molarity of given solution is 0.3004 M.

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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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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY