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 15, Problem 20PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in 100 mL of 2.25 M FeCl3 has to be calculated.

Concept Introduction:

Molarity is quantitatively defined as moles of solute in one liter of solution. For example 0.070 M AlCl3 indicates that in 1 L the moles of AlCl3 is 0.070 mol. The expression to evaluate mass is as follows:

  Mass=(Number of moles)(Molar Mass)

(a)

Expert Solution
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Explanation of Solution

FeCl3 can be broken into 1 mol Fe3+ and 2 mol Cl as follows:

  FeCl3Fe3++3Cl

Since 1 M Fe3+ is furnished by 1 M FeCl3 so molarity of Fe3+ ion due to 2.25 M FeCl3 is calculated as follows:

  Molarity of Fe3+=(2.25 M FeCl3)(1 M Fe3+1 M FeCl3)=2.25 M Fe3+

Since 2 M Cl is furnished by 1 M FeCl3 so molarity of Cl ions due to 2.25 M FeCl3 is calculated as follows:

  Molarity of Cl=(2.25 M FeCl3)(2 M Cl1 M FeCl3)=4.5 M Cl

2.25 M Fe3+ indicates that in 1000 mL moles of Fe3+ is 2.25 mol thus moles of Fe3+ in 100 mL is calculated as follows:

  Moles of Fe3+=(100 mL)(2.25 mol1000 mL)=0.225 mol

4.5 M Cl indicates that in 1000 mL the moles of Cl is 4.5 mol thus moles of Cl in 100 mL is calculated as follows:

  Moles of Cl=(100 mL)(4.5 mol1000 mL)=0.45 mol

The expression to evaluate mass is as follows:

  Mass=(Number of moles)(Molar Mass)        (1)

Substitute 0.225 mol for number of moles and 55.845 g/mol for molar mass of Fe3+ in equation (1).

  Mass=(0.225 mol)(55.845 g/mol)=12.5651 g

Substitute 0.45 mol for number of moles and 35.453 g/mol for molar mass of Cl in equation (1).

  Mass=(0.45 mol)(35.453 g/mol)=15.9538 g

Thus mass of  Fe3+ , Cl are 12.5651 g and 15.9538 g respectively.

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in 100 mL of 0.75 M NaH2PO4 has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
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Explanation of Solution

NaH2PO4 can be broken into 1 mol Na+ and  1 mol H2PO4 as follows:

  NaH2PO4Na++H2PO4

Since 1 M Na+ is furnished by 1 M NaH2PO4 so molarity of Na+ ions due to 0.75 M NaH2PO4 is calculated as follows:

  Molarity of Na+=(0.75 M NaH2PO4)(1 M Na+1 M NaH2PO4)=0.75 M Na+

Since 1 M H2PO4 is furnished by 1 M NaH2PO4 so molarity of H2PO4 ion due to 0.75 M NaH2PO4 is calculated as follows:

  Molarity of H2PO4=(0.75 M NaH2PO4)(1 M H2PO41 M NaH2PO4)=0.75 M H2PO4

0.75 M Na+ indicates that in 1000 mL the moles of Na+ is 0.75 mol thus moles of Na+ in 100 mL is calculated as follows:

  Moles of Na+=(100 mL)(0.75 mol1000 mL)=0.075 mol

0.75 M H2PO4 indicates that in 1000 mL moles of H2PO4 is 0.75 mol thus moles of H2PO4 in 100 mL is calculated as follows:

  Moles of H2PO4=(100 mL)(0.75 mol1000 mL)=0.075 mol

Substitute 0.075 mol for number of moles and 22.989 g/mol for molar mass of Na+ in equation (1).

  Mass=(0.075 mol)(22.989 g/mol)=1.7241 g

Substitute 0.075 mol for number of moles and 96.987 g/mol for molar mass of H2PO4 in equation (1).

  Mass=(0.075 mol)(96.987 g/mol)=7.274 g

Thus masses of Na+ and H2PO4 is 1.7241 g and 7.274 g respectively.

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in 100 mL of 1.20 M MgSO4 has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
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Explanation of Solution

MgSO4 can be broken into 1 mol Mg2+ and  1 mol SO42 as follows:

  MgSO4Mg2++SO42

Since 1 M Mg2+ is furnished by 1 M MgSO4 so molarity of Mg2+ ions due to 1.20 M MgSO4 is calculated as follows:

  Molarity of  Mg2+=(1.20 M MgSO4)(1 M Mg2+1 M MgSO4)=1.20 M Mg2+

Since 1 M SO42 is furnished by 1 M MgSO4 so molarity of SO42 ion due to 1.20 M MgSO4 is calculated as follows:

  Molarity of SO42=(1.20 M MgSO4)(1 M SO421 M MgSO4)=1.20 M SO42

1.20 M Mg2+ indicates that in 1000 mL the moles of Mg2+ is 1.20 mol thus moles of Mg2+ in 100 mL is calculated as follows:

  Moles of Mg2+=(100 mL)(1.20 mol1000 mL)=0.120 mol

1.20 M SO42 indicates that in 1000 mL moles of SO42 is 1.20 mol thus moles of SO42 in 100 mL is calculated as follows:

  Moles of SO42=(100 mL)(1.20 mol1000 mL)=0.120 mol

Substitute 0.120 mol for number of moles and 24.305 g/mol for molar mass of Mg2+ in equation (1).

  Mass=(0.120 mol)(24.305 g/mol)=2.9166 g

Substitute 0.120 mol for number of moles and 96.06 g/mol for molar mass of SO42 in equation (1).

  Mass=(0.120 mol)(96.06 g/mol)=11.5272 g

Thus mass of Mg2+, SO42 are 2.9166 g and 21.355 g respectively.

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in 100 mL of 0.35 M Ca(ClO3)2 has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Ca(ClO3)2 can be broken into 1 mol Ca2+ and  2 mol ClO3 as follows:

  Ca(ClO3)2Ca2++2ClO3

Since 1 M Ca2+ is furnished by 1 M Ca(ClO3)2 so molarity of Ca2+ ions due to 0.35 M Ca(ClO3)2 is calculated as follows:

  Molarity of Ca2+=(0.35 M Ca(ClO3)2)(1 M Ca2+1 M Ca(ClO3)2)=0.35 M Ca2+

Since 2 M ClO3 is furnished by 1 M Ca(ClO3)2 so molarity of ClO3 ion due to 0.35 M Ca(ClO3)2 is calculated as follows:

  Molarity of ClO3=(0.35 M Ca(ClO3)2)(2 M ClO31 M Ca(ClO3)2)=0.7 M ClO3

0.35 M Ca2+ indicates that in 1000 mL the moles of Ca2+ is 0.35 mol thus moles of Ca2+ in 100 mL is calculated as follows:

  Moles of Ca2+=(100 mL)(0.35 mol1000 mL)=0.035 mol

0.7 M ClO3 indicates that in 1000 mL moles of ClO3 is 0.7 mol thus moles of ClO3 in 100 mL is calculated as follows:

  Moles of ClO3=(100 mL)(0.7 mol1000 mL)=0.07 mol

Substitute 0.035 mol for number of moles and 40.078 g/mol for molar mass of Ca2+ in equation (1).

  Mass=(0.035 mol)(40.078 g/mol)=1.40273 g

Substitute 0.07 mol for number of moles and 99.45 g/mol for molar mass of ClO3 in equation (1).

  Mass=(0.07 mol)(99.45 g/mol)=6.9615 g

Thus mass of Ca2+ and ClO3 are 1.40273 g and 6.9615 g respectively.

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Chapter 15 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

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