Mass of each ion found in 100 mL of 1 .25 M CuBr 2 has to be calculated. Concept Introduction: Molarity is quantitatively defined as moles of solute in one liter of solution. For example, 0 .070 M AlCl 3 indicates that in 1 L the moles of AlCl 3 is 0 .070 mol . The expression to evaluate mass is as follows: Mass = ( Number of moles ) ( Molar Mass )

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 15, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $1.25 M CuBr2$ 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

Explanation of Solution

$CuBr2$ can be broken into $1 mol Cu2+$ and $2 mol Br−$ as follows:

$CuBr2⇌Cu2++2Br−$

Since $1 M Cu2+$ is furnished by $1 M CuBr2$ so molarity of $Cu2+$ ion due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Cu2+=(1.25 M CuBr2)(1 M Cu2+1 M CuBr2)=1.25 M Cu2+$

Since $2 M Br−$ is furnished by $1 M CuBr2$ so molarity of $Br−$ ions due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Br−=(1.25 M CuBr2)(2 M Br−1 M CuBr2)=2.5 M Br−$

$1.25 M Cu2+$ indicates that in $1000 mL$ the moles of $Cu2+$ is $2.5 mol$ thus moles of $Cu2+$ in $100 mL$ is calculated as follows:

$Moles of Cu2+=(100 mL)(1.25 mol1000 mL)=0.125 mol$

$2.5 M Br−$ indicates that in $1000 mL$ the moles of $Br−$ is $2.5 mol$ thus moles of $Br−$ in

$100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(2.5 mol1000 mL)=0.25 mol$

The expression to evaluate mass is as follows:

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

Substitute $0.25 mol$ for number of moles and $79.904 g/mol$ for molar mass of $Br−$ in equation (1).

$Mass=(25 mol)(79.904 g/mol)=19.976 g$

Substitute $0.125 mol$ for number of moles and $63.546 g/mol$ for molar mass of $Cu2+$ in equation (1).

$Mass=(0.125 mol)(63.546 g/mol)=7.94 g$

Thus, mass of $Cu2+$ and $Br−$ are $7.94 g$ and $19.976 g$ respectively.

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $3.50 M K3AsO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

$K3AsO4$ can be broken into $3 mol K+$ and $1 mol AsO4−$ as follows:

$K3AsO4⇌3K++AsO4−$

Since $3 M K+$ is furnished by $1 M AlCl3$ so molarity of $K+$ ions due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of K+=(3.50 M K3AsO4)(3 M K+1 M K3AsO4)=10.5 M K+$

Since $1 M AsO4−$ is furnished by $1 M K3AsO4$ so molarity of $AsO4−$ ion due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of AsO4−=(3.50 M K3AsO4)(1 M AsO4−1 M K3AsO4)=3.50 M AsO4−$

$10.5 M K+$ indicates that in $1000 mL$ the moles of $K+$ is $10.5 mol$ thus moles of $K+$ in $100 mL$ is calculated as follows:

$Moles of K+=(100 mL)(10.5 mol1000 mL)=1.05 mol$

$3.50 M AsO4−$ indicates that in $1000 mL$ moles of $AsO4−$ is $3.5 mol$ thus moles of $AsO4−$ in $100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(3.50 mol1000 mL)=0.35 mol$

Substitute $1.05 mol$ for number of moles and $39.098 g/mol$ for molar mass of $K+$ in equation (1).

$Mass=(1.05 mol)(39.098 g/mol)=41.053 g$

Substitute $0.35 mol$ for number of moles and $138.919 g/mol$ for molar mass of $AsO4−$ in equation (1).

$Mass=(0.35 mol)(138.919 g/mol)=48.621 g$

Thus, mass of $K+$ and $AsO4−$ are $41.053 g$ and $48.621 g$ respectively.

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.75 M NaHCO3$ has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

$NaHCO3$ can be broken into $1 mol Na+$ and $1 mol HCO3−$ as follows:

$NaHCO3⇌Na++HCO3−$

Since $1 M Na+$ is furnished by $1 M NaHCO3$ so molarity of $Na+$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of Na+=(0.75 M NaHCO3)(1 M Na+1 M NaHCO3)=0.75 M Na+$

Since $1 M HCO3−$ is furnished by $1 M NaHCO3$ so molarity of $HCO3−$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of HCO3−=(0.75 M NaHCO3)(1 M HCO3−1 M NaHCO3)=0.75 M HCO3−$

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

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

$0.75 M HCO3−$ indicates that in $1000 mL$ moles of $HCO3−$ is $0.75 mol$ thus moles of $HCO3−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.75 mol1000 mL)=0.075 mol$

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

$Mass=(0.075 mol)(22.989 g/mol)=1.724 g$

Substitute $0.075 mol$ for number of moles and $61.0168 g/mol$ for molar mass of $HCO3−$ in equation (1).

$Mass=(0.35 mol)(61.0168 g/mol)=21.355 g$

Thus in $100 mL$ mass of $Na+$ and $HCO3−$ are $1.724 g$ and $21.355 g$ respectively.

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.65 M (NH4)2SO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

$(NH4)2SO4$ can be broken into $2 mol NH4+$ and $1 mol SO42−$ as follows:

$(NH4)2SO4⇌2NH4++SO42−$

Since $2 M NH4+$ is furnished by $1 M (NH4)2SO4$ so molarity of $NH4+$ ions due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of NH4+=(0.65 M (NH4)2SO4)(2 M NH4+1 M (NH4)2SO4)=1.3 M NH4+$

Since $1 M SO42−$ is furnished by $1 M (NH4)2SO4$ so molarity of $SO42−$ ion due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of SO42−=(0.65 M (NH4)2SO4)(1 M SO42−1 M (NH4)2SO4)=0.65 M SO42−$

$1.3 M NH4+$ indicates that in $1000 mL$ the moles of $NH4+$ is $1.3 mol$ thus moles of $NH4+$ in $100 mL$ is calculated as follows:

$Moles of NH4+=(100 mL)(1.3 mol1000 mL)=0.13 mol$

$0.65 M SO42−$ indicates that in $1000 mL$ moles of $SO42−$ is $0.65 mol$ thus moles of $SO42−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.65 mol1000 mL)=0.065 mol$

Substitute $0.13 mol$ for number of moles and $18.039 g/mol$ for molar mass of $NH4+$ in equation (1).

$Mass=(0.13 mol)(18.039 g/mol)=2.345 g$

Substitute $0.65 mol$ for number of moles and $96.06 g/mol$ for molar mass of $SO42−$ in equation (1).

$Mass=(0.65 mol)(96.06 g/mol)=62.439 g$

Thus mass of $NH4+$ and $SO42−$ are $2.345 g$ and $62.439 g$ respectively.

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!

Students have asked these similar questions

2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled** Peak Chemical Shift (d) 5.7 1 Multiplicity multiplate .......... 5.04 double of doublet 2 4.98 double of doublet 3 4.05 doublet of quartet 4 5 LO 3.80 quartet 1.3 doublet 6 Peak Chemical Shift (d) Multiplicity

Interpreting NMR spectra is a skill that often requires some amount of practice, which, in turn, necessitates access to a collection of NMR spectra. Beyond Labz Organic Synthesis and Organic Qualitative Analysis have spectral libraries containing over 700 1H NMR spectra. In this assignment, you will take advantage of this by first predicting the NMR spectra for two closely related compounds and then checking your predictions by looking up the actual spectra in the spectra library. After completing this assignment, you may wish to select other compounds for additional practice. 1. Write the IUPAC names for the following two structures: Question 2 Question 3 2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled**

11:14 ... worksheets.beyondlabz.com 3. To check your predictions, click this link for Interpreting NMR Spectra 1. You will see a list of all the - compounds in the spectra library in alphabetical order by IUPAC name. Hovering over a name in the list will show the structure on the chalkboard. The four buttons on the top of the Spectra tab in the tray are used to select the different spectroscopic techniques for the selected compound. Make sure the NMR button has been selected. 4. Scroll through the list of names to find the names for the two compounds you have been given and click on the name to display the NMR spectrum for each. In the NMR tables below, list the chemical shift, the splitting, and the number of hydrogens associated with each peak for each compound. Compare your answers to your predictions. **Not all slots must be filled** Peak Chemical Shift (d) Multiplicity 1 2 3 4 5

Answer 2

Question

Chapter 15, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $1.25 M CuBr2$ 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

Explanation of Solution

$CuBr2$ can be broken into $1 mol Cu2+$ and $2 mol Br−$ as follows:

$CuBr2⇌Cu2++2Br−$

Since $1 M Cu2+$ is furnished by $1 M CuBr2$ so molarity of $Cu2+$ ion due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Cu2+=(1.25 M CuBr2)(1 M Cu2+1 M CuBr2)=1.25 M Cu2+$

Since $2 M Br−$ is furnished by $1 M CuBr2$ so molarity of $Br−$ ions due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Br−=(1.25 M CuBr2)(2 M Br−1 M CuBr2)=2.5 M Br−$

$1.25 M Cu2+$ indicates that in $1000 mL$ the moles of $Cu2+$ is $2.5 mol$ thus moles of $Cu2+$ in $100 mL$ is calculated as follows:

$Moles of Cu2+=(100 mL)(1.25 mol1000 mL)=0.125 mol$

$2.5 M Br−$ indicates that in $1000 mL$ the moles of $Br−$ is $2.5 mol$ thus moles of $Br−$ in

$100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(2.5 mol1000 mL)=0.25 mol$

The expression to evaluate mass is as follows:

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

Substitute $0.25 mol$ for number of moles and $79.904 g/mol$ for molar mass of $Br−$ in equation (1).

$Mass=(25 mol)(79.904 g/mol)=19.976 g$

Substitute $0.125 mol$ for number of moles and $63.546 g/mol$ for molar mass of $Cu2+$ in equation (1).

$Mass=(0.125 mol)(63.546 g/mol)=7.94 g$

Thus, mass of $Cu2+$ and $Br−$ are $7.94 g$ and $19.976 g$ respectively.

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $3.50 M K3AsO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

$K3AsO4$ can be broken into $3 mol K+$ and $1 mol AsO4−$ as follows:

$K3AsO4⇌3K++AsO4−$

Since $3 M K+$ is furnished by $1 M AlCl3$ so molarity of $K+$ ions due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of K+=(3.50 M K3AsO4)(3 M K+1 M K3AsO4)=10.5 M K+$

Since $1 M AsO4−$ is furnished by $1 M K3AsO4$ so molarity of $AsO4−$ ion due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of AsO4−=(3.50 M K3AsO4)(1 M AsO4−1 M K3AsO4)=3.50 M AsO4−$

$10.5 M K+$ indicates that in $1000 mL$ the moles of $K+$ is $10.5 mol$ thus moles of $K+$ in $100 mL$ is calculated as follows:

$Moles of K+=(100 mL)(10.5 mol1000 mL)=1.05 mol$

$3.50 M AsO4−$ indicates that in $1000 mL$ moles of $AsO4−$ is $3.5 mol$ thus moles of $AsO4−$ in $100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(3.50 mol1000 mL)=0.35 mol$

Substitute $1.05 mol$ for number of moles and $39.098 g/mol$ for molar mass of $K+$ in equation (1).

$Mass=(1.05 mol)(39.098 g/mol)=41.053 g$

Substitute $0.35 mol$ for number of moles and $138.919 g/mol$ for molar mass of $AsO4−$ in equation (1).

$Mass=(0.35 mol)(138.919 g/mol)=48.621 g$

Thus, mass of $K+$ and $AsO4−$ are $41.053 g$ and $48.621 g$ respectively.

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.75 M NaHCO3$ has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

$NaHCO3$ can be broken into $1 mol Na+$ and $1 mol HCO3−$ as follows:

$NaHCO3⇌Na++HCO3−$

Since $1 M Na+$ is furnished by $1 M NaHCO3$ so molarity of $Na+$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of Na+=(0.75 M NaHCO3)(1 M Na+1 M NaHCO3)=0.75 M Na+$

Since $1 M HCO3−$ is furnished by $1 M NaHCO3$ so molarity of $HCO3−$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of HCO3−=(0.75 M NaHCO3)(1 M HCO3−1 M NaHCO3)=0.75 M HCO3−$

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

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

$0.75 M HCO3−$ indicates that in $1000 mL$ moles of $HCO3−$ is $0.75 mol$ thus moles of $HCO3−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.75 mol1000 mL)=0.075 mol$

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

$Mass=(0.075 mol)(22.989 g/mol)=1.724 g$

Substitute $0.075 mol$ for number of moles and $61.0168 g/mol$ for molar mass of $HCO3−$ in equation (1).

$Mass=(0.35 mol)(61.0168 g/mol)=21.355 g$

Thus in $100 mL$ mass of $Na+$ and $HCO3−$ are $1.724 g$ and $21.355 g$ respectively.

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.65 M (NH4)2SO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

$(NH4)2SO4$ can be broken into $2 mol NH4+$ and $1 mol SO42−$ as follows:

$(NH4)2SO4⇌2NH4++SO42−$

Since $2 M NH4+$ is furnished by $1 M (NH4)2SO4$ so molarity of $NH4+$ ions due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of NH4+=(0.65 M (NH4)2SO4)(2 M NH4+1 M (NH4)2SO4)=1.3 M NH4+$

Since $1 M SO42−$ is furnished by $1 M (NH4)2SO4$ so molarity of $SO42−$ ion due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of SO42−=(0.65 M (NH4)2SO4)(1 M SO42−1 M (NH4)2SO4)=0.65 M SO42−$

$1.3 M NH4+$ indicates that in $1000 mL$ the moles of $NH4+$ is $1.3 mol$ thus moles of $NH4+$ in $100 mL$ is calculated as follows:

$Moles of NH4+=(100 mL)(1.3 mol1000 mL)=0.13 mol$

$0.65 M SO42−$ indicates that in $1000 mL$ moles of $SO42−$ is $0.65 mol$ thus moles of $SO42−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.65 mol1000 mL)=0.065 mol$

Substitute $0.13 mol$ for number of moles and $18.039 g/mol$ for molar mass of $NH4+$ in equation (1).

$Mass=(0.13 mol)(18.039 g/mol)=2.345 g$

Substitute $0.65 mol$ for number of moles and $96.06 g/mol$ for molar mass of $SO42−$ in equation (1).

$Mass=(0.65 mol)(96.06 g/mol)=62.439 g$

Thus mass of $NH4+$ and $SO42−$ are $2.345 g$ and $62.439 g$ respectively.

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!

Answer 3

Question

Chapter 15, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $1.25 M CuBr2$ 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

Explanation of Solution

$CuBr2$ can be broken into $1 mol Cu2+$ and $2 mol Br−$ as follows:

$CuBr2⇌Cu2++2Br−$

Since $1 M Cu2+$ is furnished by $1 M CuBr2$ so molarity of $Cu2+$ ion due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Cu2+=(1.25 M CuBr2)(1 M Cu2+1 M CuBr2)=1.25 M Cu2+$

Since $2 M Br−$ is furnished by $1 M CuBr2$ so molarity of $Br−$ ions due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Br−=(1.25 M CuBr2)(2 M Br−1 M CuBr2)=2.5 M Br−$

$1.25 M Cu2+$ indicates that in $1000 mL$ the moles of $Cu2+$ is $2.5 mol$ thus moles of $Cu2+$ in $100 mL$ is calculated as follows:

$Moles of Cu2+=(100 mL)(1.25 mol1000 mL)=0.125 mol$

$2.5 M Br−$ indicates that in $1000 mL$ the moles of $Br−$ is $2.5 mol$ thus moles of $Br−$ in

$100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(2.5 mol1000 mL)=0.25 mol$

The expression to evaluate mass is as follows:

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

Substitute $0.25 mol$ for number of moles and $79.904 g/mol$ for molar mass of $Br−$ in equation (1).

$Mass=(25 mol)(79.904 g/mol)=19.976 g$

Substitute $0.125 mol$ for number of moles and $63.546 g/mol$ for molar mass of $Cu2+$ in equation (1).

$Mass=(0.125 mol)(63.546 g/mol)=7.94 g$

Thus, mass of $Cu2+$ and $Br−$ are $7.94 g$ and $19.976 g$ respectively.

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $3.50 M K3AsO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

$K3AsO4$ can be broken into $3 mol K+$ and $1 mol AsO4−$ as follows:

$K3AsO4⇌3K++AsO4−$

Since $3 M K+$ is furnished by $1 M AlCl3$ so molarity of $K+$ ions due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of K+=(3.50 M K3AsO4)(3 M K+1 M K3AsO4)=10.5 M K+$

Since $1 M AsO4−$ is furnished by $1 M K3AsO4$ so molarity of $AsO4−$ ion due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of AsO4−=(3.50 M K3AsO4)(1 M AsO4−1 M K3AsO4)=3.50 M AsO4−$

$10.5 M K+$ indicates that in $1000 mL$ the moles of $K+$ is $10.5 mol$ thus moles of $K+$ in $100 mL$ is calculated as follows:

$Moles of K+=(100 mL)(10.5 mol1000 mL)=1.05 mol$

$3.50 M AsO4−$ indicates that in $1000 mL$ moles of $AsO4−$ is $3.5 mol$ thus moles of $AsO4−$ in $100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(3.50 mol1000 mL)=0.35 mol$

Substitute $1.05 mol$ for number of moles and $39.098 g/mol$ for molar mass of $K+$ in equation (1).

$Mass=(1.05 mol)(39.098 g/mol)=41.053 g$

Substitute $0.35 mol$ for number of moles and $138.919 g/mol$ for molar mass of $AsO4−$ in equation (1).

$Mass=(0.35 mol)(138.919 g/mol)=48.621 g$

Thus, mass of $K+$ and $AsO4−$ are $41.053 g$ and $48.621 g$ respectively.

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.75 M NaHCO3$ has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

$NaHCO3$ can be broken into $1 mol Na+$ and $1 mol HCO3−$ as follows:

$NaHCO3⇌Na++HCO3−$

Since $1 M Na+$ is furnished by $1 M NaHCO3$ so molarity of $Na+$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of Na+=(0.75 M NaHCO3)(1 M Na+1 M NaHCO3)=0.75 M Na+$

Since $1 M HCO3−$ is furnished by $1 M NaHCO3$ so molarity of $HCO3−$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of HCO3−=(0.75 M NaHCO3)(1 M HCO3−1 M NaHCO3)=0.75 M HCO3−$

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

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

$0.75 M HCO3−$ indicates that in $1000 mL$ moles of $HCO3−$ is $0.75 mol$ thus moles of $HCO3−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.75 mol1000 mL)=0.075 mol$

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

$Mass=(0.075 mol)(22.989 g/mol)=1.724 g$

Substitute $0.075 mol$ for number of moles and $61.0168 g/mol$ for molar mass of $HCO3−$ in equation (1).

$Mass=(0.35 mol)(61.0168 g/mol)=21.355 g$

Thus in $100 mL$ mass of $Na+$ and $HCO3−$ are $1.724 g$ and $21.355 g$ respectively.

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.65 M (NH4)2SO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

$(NH4)2SO4$ can be broken into $2 mol NH4+$ and $1 mol SO42−$ as follows:

$(NH4)2SO4⇌2NH4++SO42−$

Since $2 M NH4+$ is furnished by $1 M (NH4)2SO4$ so molarity of $NH4+$ ions due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of NH4+=(0.65 M (NH4)2SO4)(2 M NH4+1 M (NH4)2SO4)=1.3 M NH4+$

Since $1 M SO42−$ is furnished by $1 M (NH4)2SO4$ so molarity of $SO42−$ ion due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of SO42−=(0.65 M (NH4)2SO4)(1 M SO42−1 M (NH4)2SO4)=0.65 M SO42−$

$1.3 M NH4+$ indicates that in $1000 mL$ the moles of $NH4+$ is $1.3 mol$ thus moles of $NH4+$ in $100 mL$ is calculated as follows:

$Moles of NH4+=(100 mL)(1.3 mol1000 mL)=0.13 mol$

$0.65 M SO42−$ indicates that in $1000 mL$ moles of $SO42−$ is $0.65 mol$ thus moles of $SO42−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.65 mol1000 mL)=0.065 mol$

Substitute $0.13 mol$ for number of moles and $18.039 g/mol$ for molar mass of $NH4+$ in equation (1).

$Mass=(0.13 mol)(18.039 g/mol)=2.345 g$

Substitute $0.65 mol$ for number of moles and $96.06 g/mol$ for molar mass of $SO42−$ in equation (1).

$Mass=(0.65 mol)(96.06 g/mol)=62.439 g$

Thus mass of $NH4+$ and $SO42−$ are $2.345 g$ and $62.439 g$ respectively.

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!

Answer 4

Question

Chapter 15, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $1.25 M CuBr2$ 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

Explanation of Solution

$CuBr2$ can be broken into $1 mol Cu2+$ and $2 mol Br−$ as follows:

$CuBr2⇌Cu2++2Br−$

Since $1 M Cu2+$ is furnished by $1 M CuBr2$ so molarity of $Cu2+$ ion due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Cu2+=(1.25 M CuBr2)(1 M Cu2+1 M CuBr2)=1.25 M Cu2+$

Since $2 M Br−$ is furnished by $1 M CuBr2$ so molarity of $Br−$ ions due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Br−=(1.25 M CuBr2)(2 M Br−1 M CuBr2)=2.5 M Br−$

$1.25 M Cu2+$ indicates that in $1000 mL$ the moles of $Cu2+$ is $2.5 mol$ thus moles of $Cu2+$ in $100 mL$ is calculated as follows:

$Moles of Cu2+=(100 mL)(1.25 mol1000 mL)=0.125 mol$

$2.5 M Br−$ indicates that in $1000 mL$ the moles of $Br−$ is $2.5 mol$ thus moles of $Br−$ in

$100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(2.5 mol1000 mL)=0.25 mol$

The expression to evaluate mass is as follows:

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

Substitute $0.25 mol$ for number of moles and $79.904 g/mol$ for molar mass of $Br−$ in equation (1).

$Mass=(25 mol)(79.904 g/mol)=19.976 g$

Substitute $0.125 mol$ for number of moles and $63.546 g/mol$ for molar mass of $Cu2+$ in equation (1).

$Mass=(0.125 mol)(63.546 g/mol)=7.94 g$

Thus, mass of $Cu2+$ and $Br−$ are $7.94 g$ and $19.976 g$ respectively.

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $3.50 M K3AsO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

$K3AsO4$ can be broken into $3 mol K+$ and $1 mol AsO4−$ as follows:

$K3AsO4⇌3K++AsO4−$

Since $3 M K+$ is furnished by $1 M AlCl3$ so molarity of $K+$ ions due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of K+=(3.50 M K3AsO4)(3 M K+1 M K3AsO4)=10.5 M K+$

Since $1 M AsO4−$ is furnished by $1 M K3AsO4$ so molarity of $AsO4−$ ion due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of AsO4−=(3.50 M K3AsO4)(1 M AsO4−1 M K3AsO4)=3.50 M AsO4−$

$10.5 M K+$ indicates that in $1000 mL$ the moles of $K+$ is $10.5 mol$ thus moles of $K+$ in $100 mL$ is calculated as follows:

$Moles of K+=(100 mL)(10.5 mol1000 mL)=1.05 mol$

$3.50 M AsO4−$ indicates that in $1000 mL$ moles of $AsO4−$ is $3.5 mol$ thus moles of $AsO4−$ in $100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(3.50 mol1000 mL)=0.35 mol$

Substitute $1.05 mol$ for number of moles and $39.098 g/mol$ for molar mass of $K+$ in equation (1).

$Mass=(1.05 mol)(39.098 g/mol)=41.053 g$

Substitute $0.35 mol$ for number of moles and $138.919 g/mol$ for molar mass of $AsO4−$ in equation (1).

$Mass=(0.35 mol)(138.919 g/mol)=48.621 g$

Thus, mass of $K+$ and $AsO4−$ are $41.053 g$ and $48.621 g$ respectively.

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.75 M NaHCO3$ has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

$NaHCO3$ can be broken into $1 mol Na+$ and $1 mol HCO3−$ as follows:

$NaHCO3⇌Na++HCO3−$

Since $1 M Na+$ is furnished by $1 M NaHCO3$ so molarity of $Na+$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of Na+=(0.75 M NaHCO3)(1 M Na+1 M NaHCO3)=0.75 M Na+$

Since $1 M HCO3−$ is furnished by $1 M NaHCO3$ so molarity of $HCO3−$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of HCO3−=(0.75 M NaHCO3)(1 M HCO3−1 M NaHCO3)=0.75 M HCO3−$

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

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

$0.75 M HCO3−$ indicates that in $1000 mL$ moles of $HCO3−$ is $0.75 mol$ thus moles of $HCO3−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.75 mol1000 mL)=0.075 mol$

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

$Mass=(0.075 mol)(22.989 g/mol)=1.724 g$

Substitute $0.075 mol$ for number of moles and $61.0168 g/mol$ for molar mass of $HCO3−$ in equation (1).

$Mass=(0.35 mol)(61.0168 g/mol)=21.355 g$

Thus in $100 mL$ mass of $Na+$ and $HCO3−$ are $1.724 g$ and $21.355 g$ respectively.

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.65 M (NH4)2SO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

$(NH4)2SO4$ can be broken into $2 mol NH4+$ and $1 mol SO42−$ as follows:

$(NH4)2SO4⇌2NH4++SO42−$

Since $2 M NH4+$ is furnished by $1 M (NH4)2SO4$ so molarity of $NH4+$ ions due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of NH4+=(0.65 M (NH4)2SO4)(2 M NH4+1 M (NH4)2SO4)=1.3 M NH4+$

Since $1 M SO42−$ is furnished by $1 M (NH4)2SO4$ so molarity of $SO42−$ ion due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of SO42−=(0.65 M (NH4)2SO4)(1 M SO42−1 M (NH4)2SO4)=0.65 M SO42−$

$1.3 M NH4+$ indicates that in $1000 mL$ the moles of $NH4+$ is $1.3 mol$ thus moles of $NH4+$ in $100 mL$ is calculated as follows:

$Moles of NH4+=(100 mL)(1.3 mol1000 mL)=0.13 mol$

$0.65 M SO42−$ indicates that in $1000 mL$ moles of $SO42−$ is $0.65 mol$ thus moles of $SO42−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.65 mol1000 mL)=0.065 mol$

Substitute $0.13 mol$ for number of moles and $18.039 g/mol$ for molar mass of $NH4+$ in equation (1).

$Mass=(0.13 mol)(18.039 g/mol)=2.345 g$

Substitute $0.65 mol$ for number of moles and $96.06 g/mol$ for molar mass of $SO42−$ in equation (1).

$Mass=(0.65 mol)(96.06 g/mol)=62.439 g$

Thus mass of $NH4+$ and $SO42−$ are $2.345 g$ and $62.439 g$ respectively.

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!

Answer 5

Chapter 15, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $1.25 M CuBr2$ 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

Explanation of Solution

$CuBr2$ can be broken into $1 mol Cu2+$ and $2 mol Br−$ as follows:

$CuBr2⇌Cu2++2Br−$

Since $1 M Cu2+$ is furnished by $1 M CuBr2$ so molarity of $Cu2+$ ion due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Cu2+=(1.25 M CuBr2)(1 M Cu2+1 M CuBr2)=1.25 M Cu2+$

Since $2 M Br−$ is furnished by $1 M CuBr2$ so molarity of $Br−$ ions due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Br−=(1.25 M CuBr2)(2 M Br−1 M CuBr2)=2.5 M Br−$

$1.25 M Cu2+$ indicates that in $1000 mL$ the moles of $Cu2+$ is $2.5 mol$ thus moles of $Cu2+$ in $100 mL$ is calculated as follows:

$Moles of Cu2+=(100 mL)(1.25 mol1000 mL)=0.125 mol$

$2.5 M Br−$ indicates that in $1000 mL$ the moles of $Br−$ is $2.5 mol$ thus moles of $Br−$ in

$100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(2.5 mol1000 mL)=0.25 mol$

The expression to evaluate mass is as follows:

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

Substitute $0.25 mol$ for number of moles and $79.904 g/mol$ for molar mass of $Br−$ in equation (1).

$Mass=(25 mol)(79.904 g/mol)=19.976 g$

Substitute $0.125 mol$ for number of moles and $63.546 g/mol$ for molar mass of $Cu2+$ in equation (1).

$Mass=(0.125 mol)(63.546 g/mol)=7.94 g$

Thus, mass of $Cu2+$ and $Br−$ are $7.94 g$ and $19.976 g$ respectively.

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $3.50 M K3AsO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

$K3AsO4$ can be broken into $3 mol K+$ and $1 mol AsO4−$ as follows:

$K3AsO4⇌3K++AsO4−$

Since $3 M K+$ is furnished by $1 M AlCl3$ so molarity of $K+$ ions due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of K+=(3.50 M K3AsO4)(3 M K+1 M K3AsO4)=10.5 M K+$

Since $1 M AsO4−$ is furnished by $1 M K3AsO4$ so molarity of $AsO4−$ ion due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of AsO4−=(3.50 M K3AsO4)(1 M AsO4−1 M K3AsO4)=3.50 M AsO4−$

$10.5 M K+$ indicates that in $1000 mL$ the moles of $K+$ is $10.5 mol$ thus moles of $K+$ in $100 mL$ is calculated as follows:

$Moles of K+=(100 mL)(10.5 mol1000 mL)=1.05 mol$

$3.50 M AsO4−$ indicates that in $1000 mL$ moles of $AsO4−$ is $3.5 mol$ thus moles of $AsO4−$ in $100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(3.50 mol1000 mL)=0.35 mol$

Substitute $1.05 mol$ for number of moles and $39.098 g/mol$ for molar mass of $K+$ in equation (1).

$Mass=(1.05 mol)(39.098 g/mol)=41.053 g$

Substitute $0.35 mol$ for number of moles and $138.919 g/mol$ for molar mass of $AsO4−$ in equation (1).

$Mass=(0.35 mol)(138.919 g/mol)=48.621 g$

Thus, mass of $K+$ and $AsO4−$ are $41.053 g$ and $48.621 g$ respectively.

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.75 M NaHCO3$ has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

$NaHCO3$ can be broken into $1 mol Na+$ and $1 mol HCO3−$ as follows:

$NaHCO3⇌Na++HCO3−$

Since $1 M Na+$ is furnished by $1 M NaHCO3$ so molarity of $Na+$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of Na+=(0.75 M NaHCO3)(1 M Na+1 M NaHCO3)=0.75 M Na+$

Since $1 M HCO3−$ is furnished by $1 M NaHCO3$ so molarity of $HCO3−$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of HCO3−=(0.75 M NaHCO3)(1 M HCO3−1 M NaHCO3)=0.75 M HCO3−$

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

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

$0.75 M HCO3−$ indicates that in $1000 mL$ moles of $HCO3−$ is $0.75 mol$ thus moles of $HCO3−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.75 mol1000 mL)=0.075 mol$

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

$Mass=(0.075 mol)(22.989 g/mol)=1.724 g$

Substitute $0.075 mol$ for number of moles and $61.0168 g/mol$ for molar mass of $HCO3−$ in equation (1).

$Mass=(0.35 mol)(61.0168 g/mol)=21.355 g$

Thus in $100 mL$ mass of $Na+$ and $HCO3−$ are $1.724 g$ and $21.355 g$ respectively.

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.65 M (NH4)2SO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

$(NH4)2SO4$ can be broken into $2 mol NH4+$ and $1 mol SO42−$ as follows:

$(NH4)2SO4⇌2NH4++SO42−$

Since $2 M NH4+$ is furnished by $1 M (NH4)2SO4$ so molarity of $NH4+$ ions due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of NH4+=(0.65 M (NH4)2SO4)(2 M NH4+1 M (NH4)2SO4)=1.3 M NH4+$

Since $1 M SO42−$ is furnished by $1 M (NH4)2SO4$ so molarity of $SO42−$ ion due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of SO42−=(0.65 M (NH4)2SO4)(1 M SO42−1 M (NH4)2SO4)=0.65 M SO42−$

$1.3 M NH4+$ indicates that in $1000 mL$ the moles of $NH4+$ is $1.3 mol$ thus moles of $NH4+$ in $100 mL$ is calculated as follows:

$Moles of NH4+=(100 mL)(1.3 mol1000 mL)=0.13 mol$

$0.65 M SO42−$ indicates that in $1000 mL$ moles of $SO42−$ is $0.65 mol$ thus moles of $SO42−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.65 mol1000 mL)=0.065 mol$

Substitute $0.13 mol$ for number of moles and $18.039 g/mol$ for molar mass of $NH4+$ in equation (1).

$Mass=(0.13 mol)(18.039 g/mol)=2.345 g$

Substitute $0.65 mol$ for number of moles and $96.06 g/mol$ for molar mass of $SO42−$ in equation (1).

$Mass=(0.65 mol)(96.06 g/mol)=62.439 g$

Thus mass of $NH4+$ and $SO42−$ are $2.345 g$ and $62.439 g$ respectively.

Answer 6

Chapter 15, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $1.25 M CuBr2$ 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

Explanation of Solution

$CuBr2$ can be broken into $1 mol Cu2+$ and $2 mol Br−$ as follows:

$CuBr2⇌Cu2++2Br−$

Since $1 M Cu2+$ is furnished by $1 M CuBr2$ so molarity of $Cu2+$ ion due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Cu2+=(1.25 M CuBr2)(1 M Cu2+1 M CuBr2)=1.25 M Cu2+$

Since $2 M Br−$ is furnished by $1 M CuBr2$ so molarity of $Br−$ ions due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Br−=(1.25 M CuBr2)(2 M Br−1 M CuBr2)=2.5 M Br−$

$1.25 M Cu2+$ indicates that in $1000 mL$ the moles of $Cu2+$ is $2.5 mol$ thus moles of $Cu2+$ in $100 mL$ is calculated as follows:

$Moles of Cu2+=(100 mL)(1.25 mol1000 mL)=0.125 mol$

$2.5 M Br−$ indicates that in $1000 mL$ the moles of $Br−$ is $2.5 mol$ thus moles of $Br−$ in

$100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(2.5 mol1000 mL)=0.25 mol$

The expression to evaluate mass is as follows:

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

Substitute $0.25 mol$ for number of moles and $79.904 g/mol$ for molar mass of $Br−$ in equation (1).

$Mass=(25 mol)(79.904 g/mol)=19.976 g$

Substitute $0.125 mol$ for number of moles and $63.546 g/mol$ for molar mass of $Cu2+$ in equation (1).

$Mass=(0.125 mol)(63.546 g/mol)=7.94 g$

Thus, mass of $Cu2+$ and $Br−$ are $7.94 g$ and $19.976 g$ respectively.

Answer 7

Chapter 15, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $1.25 M CuBr2$ 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:

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

Answer 8

(a)

Expert Solution

Explanation of Solution

$CuBr2$ can be broken into $1 mol Cu2+$ and $2 mol Br−$ as follows:

$CuBr2⇌Cu2++2Br−$

Since $1 M Cu2+$ is furnished by $1 M CuBr2$ so molarity of $Cu2+$ ion due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Cu2+=(1.25 M CuBr2)(1 M Cu2+1 M CuBr2)=1.25 M Cu2+$

Since $2 M Br−$ is furnished by $1 M CuBr2$ so molarity of $Br−$ ions due to $1.25 M CuBr2$ is calculated as follows:

$Molarity of Br−=(1.25 M CuBr2)(2 M Br−1 M CuBr2)=2.5 M Br−$

$1.25 M Cu2+$ indicates that in $1000 mL$ the moles of $Cu2+$ is $2.5 mol$ thus moles of $Cu2+$ in $100 mL$ is calculated as follows:

$Moles of Cu2+=(100 mL)(1.25 mol1000 mL)=0.125 mol$

$2.5 M Br−$ indicates that in $1000 mL$ the moles of $Br−$ is $2.5 mol$ thus moles of $Br−$ in

$100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(2.5 mol1000 mL)=0.25 mol$

The expression to evaluate mass is as follows:

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

Substitute $0.25 mol$ for number of moles and $79.904 g/mol$ for molar mass of $Br−$ in equation (1).

$Mass=(25 mol)(79.904 g/mol)=19.976 g$

Substitute $0.125 mol$ for number of moles and $63.546 g/mol$ for molar mass of $Cu2+$ in equation (1).

$Mass=(0.125 mol)(63.546 g/mol)=7.94 g$

Thus, mass of $Cu2+$ and $Br−$ are $7.94 g$ and $19.976 g$ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $3.50 M K3AsO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

$K3AsO4$ can be broken into $3 mol K+$ and $1 mol AsO4−$ as follows:

$K3AsO4⇌3K++AsO4−$

Since $3 M K+$ is furnished by $1 M AlCl3$ so molarity of $K+$ ions due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of K+=(3.50 M K3AsO4)(3 M K+1 M K3AsO4)=10.5 M K+$

Since $1 M AsO4−$ is furnished by $1 M K3AsO4$ so molarity of $AsO4−$ ion due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of AsO4−=(3.50 M K3AsO4)(1 M AsO4−1 M K3AsO4)=3.50 M AsO4−$

$10.5 M K+$ indicates that in $1000 mL$ the moles of $K+$ is $10.5 mol$ thus moles of $K+$ in $100 mL$ is calculated as follows:

$Moles of K+=(100 mL)(10.5 mol1000 mL)=1.05 mol$

$3.50 M AsO4−$ indicates that in $1000 mL$ moles of $AsO4−$ is $3.5 mol$ thus moles of $AsO4−$ in $100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(3.50 mol1000 mL)=0.35 mol$

Substitute $1.05 mol$ for number of moles and $39.098 g/mol$ for molar mass of $K+$ in equation (1).

$Mass=(1.05 mol)(39.098 g/mol)=41.053 g$

Substitute $0.35 mol$ for number of moles and $138.919 g/mol$ for molar mass of $AsO4−$ in equation (1).

$Mass=(0.35 mol)(138.919 g/mol)=48.621 g$

Thus, mass of $K+$ and $AsO4−$ are $41.053 g$ and $48.621 g$ respectively.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $3.50 M K3AsO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 14

(b)

Expert Solution

Explanation of Solution

$K3AsO4$ can be broken into $3 mol K+$ and $1 mol AsO4−$ as follows:

$K3AsO4⇌3K++AsO4−$

Since $3 M K+$ is furnished by $1 M AlCl3$ so molarity of $K+$ ions due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of K+=(3.50 M K3AsO4)(3 M K+1 M K3AsO4)=10.5 M K+$

Since $1 M AsO4−$ is furnished by $1 M K3AsO4$ so molarity of $AsO4−$ ion due to $3.50 M K3AsO4$ is calculated as follows:

$Molarity of AsO4−=(3.50 M K3AsO4)(1 M AsO4−1 M K3AsO4)=3.50 M AsO4−$

$10.5 M K+$ indicates that in $1000 mL$ the moles of $K+$ is $10.5 mol$ thus moles of $K+$ in $100 mL$ is calculated as follows:

$Moles of K+=(100 mL)(10.5 mol1000 mL)=1.05 mol$

$3.50 M AsO4−$ indicates that in $1000 mL$ moles of $AsO4−$ is $3.5 mol$ thus moles of $AsO4−$ in $100 mL$ is calculated as follows:

$Moles of Br−=(100 mL)(3.50 mol1000 mL)=0.35 mol$

Substitute $1.05 mol$ for number of moles and $39.098 g/mol$ for molar mass of $K+$ in equation (1).

$Mass=(1.05 mol)(39.098 g/mol)=41.053 g$

Substitute $0.35 mol$ for number of moles and $138.919 g/mol$ for molar mass of $AsO4−$ in equation (1).

$Mass=(0.35 mol)(138.919 g/mol)=48.621 g$

Thus, mass of $K+$ and $AsO4−$ are $41.053 g$ and $48.621 g$ respectively.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.75 M NaHCO3$ has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

$NaHCO3$ can be broken into $1 mol Na+$ and $1 mol HCO3−$ as follows:

$NaHCO3⇌Na++HCO3−$

Since $1 M Na+$ is furnished by $1 M NaHCO3$ so molarity of $Na+$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of Na+=(0.75 M NaHCO3)(1 M Na+1 M NaHCO3)=0.75 M Na+$

Since $1 M HCO3−$ is furnished by $1 M NaHCO3$ so molarity of $HCO3−$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of HCO3−=(0.75 M NaHCO3)(1 M HCO3−1 M NaHCO3)=0.75 M HCO3−$

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

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

$0.75 M HCO3−$ indicates that in $1000 mL$ moles of $HCO3−$ is $0.75 mol$ thus moles of $HCO3−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.75 mol1000 mL)=0.075 mol$

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

$Mass=(0.075 mol)(22.989 g/mol)=1.724 g$

Substitute $0.075 mol$ for number of moles and $61.0168 g/mol$ for molar mass of $HCO3−$ in equation (1).

$Mass=(0.35 mol)(61.0168 g/mol)=21.355 g$

Thus in $100 mL$ mass of $Na+$ and $HCO3−$ are $1.724 g$ and $21.355 g$ respectively.

Answer 19

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.75 M NaHCO3$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 20

(c)

Expert Solution

Explanation of Solution

$NaHCO3$ can be broken into $1 mol Na+$ and $1 mol HCO3−$ as follows:

$NaHCO3⇌Na++HCO3−$

Since $1 M Na+$ is furnished by $1 M NaHCO3$ so molarity of $Na+$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of Na+=(0.75 M NaHCO3)(1 M Na+1 M NaHCO3)=0.75 M Na+$

Since $1 M HCO3−$ is furnished by $1 M NaHCO3$ so molarity of $HCO3−$ ions due to $0.75 M NaHCO3$ is calculated as follows:

$Molarity of HCO3−=(0.75 M NaHCO3)(1 M HCO3−1 M NaHCO3)=0.75 M HCO3−$

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

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

$0.75 M HCO3−$ indicates that in $1000 mL$ moles of $HCO3−$ is $0.75 mol$ thus moles of $HCO3−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.75 mol1000 mL)=0.075 mol$

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

$Mass=(0.075 mol)(22.989 g/mol)=1.724 g$

Substitute $0.075 mol$ for number of moles and $61.0168 g/mol$ for molar mass of $HCO3−$ in equation (1).

$Mass=(0.35 mol)(61.0168 g/mol)=21.355 g$

Thus in $100 mL$ mass of $Na+$ and $HCO3−$ are $1.724 g$ and $21.355 g$ respectively.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.65 M (NH4)2SO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

$(NH4)2SO4$ can be broken into $2 mol NH4+$ and $1 mol SO42−$ as follows:

$(NH4)2SO4⇌2NH4++SO42−$

Since $2 M NH4+$ is furnished by $1 M (NH4)2SO4$ so molarity of $NH4+$ ions due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of NH4+=(0.65 M (NH4)2SO4)(2 M NH4+1 M (NH4)2SO4)=1.3 M NH4+$

Since $1 M SO42−$ is furnished by $1 M (NH4)2SO4$ so molarity of $SO42−$ ion due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of SO42−=(0.65 M (NH4)2SO4)(1 M SO42−1 M (NH4)2SO4)=0.65 M SO42−$

$1.3 M NH4+$ indicates that in $1000 mL$ the moles of $NH4+$ is $1.3 mol$ thus moles of $NH4+$ in $100 mL$ is calculated as follows:

$Moles of NH4+=(100 mL)(1.3 mol1000 mL)=0.13 mol$

$0.65 M SO42−$ indicates that in $1000 mL$ moles of $SO42−$ is $0.65 mol$ thus moles of $SO42−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.65 mol1000 mL)=0.065 mol$

Substitute $0.13 mol$ for number of moles and $18.039 g/mol$ for molar mass of $NH4+$ in equation (1).

$Mass=(0.13 mol)(18.039 g/mol)=2.345 g$

Substitute $0.65 mol$ for number of moles and $96.06 g/mol$ for molar mass of $SO42−$ in equation (1).

$Mass=(0.65 mol)(96.06 g/mol)=62.439 g$

Thus mass of $NH4+$ and $SO42−$ are $2.345 g$ and $62.439 g$ respectively.

Answer 25

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100 mL$ of $0.65 M (NH4)2SO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 26

(d)

Expert Solution

Explanation of Solution

$(NH4)2SO4$ can be broken into $2 mol NH4+$ and $1 mol SO42−$ as follows:

$(NH4)2SO4⇌2NH4++SO42−$

Since $2 M NH4+$ is furnished by $1 M (NH4)2SO4$ so molarity of $NH4+$ ions due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of NH4+=(0.65 M (NH4)2SO4)(2 M NH4+1 M (NH4)2SO4)=1.3 M NH4+$

Since $1 M SO42−$ is furnished by $1 M (NH4)2SO4$ so molarity of $SO42−$ ion due to $0.65 M (NH4)2SO4$ is calculated as follows:

$Molarity of SO42−=(0.65 M (NH4)2SO4)(1 M SO42−1 M (NH4)2SO4)=0.65 M SO42−$

$1.3 M NH4+$ indicates that in $1000 mL$ the moles of $NH4+$ is $1.3 mol$ thus moles of $NH4+$ in $100 mL$ is calculated as follows:

$Moles of NH4+=(100 mL)(1.3 mol1000 mL)=0.13 mol$

$0.65 M SO42−$ indicates that in $1000 mL$ moles of $SO42−$ is $0.65 mol$ thus moles of $SO42−$ in $100 mL$ is calculated as follows:

$Moles of HCO3−=(100 mL)(0.65 mol1000 mL)=0.065 mol$

Substitute $0.13 mol$ for number of moles and $18.039 g/mol$ for molar mass of $NH4+$ in equation (1).

$Mass=(0.13 mol)(18.039 g/mol)=2.345 g$

Substitute $0.65 mol$ for number of moles and $96.06 g/mol$ for molar mass of $SO42−$ in equation (1).

$Mass=(0.65 mol)(96.06 g/mol)=62.439 g$

Thus mass of $NH4+$ and $SO42−$ are $2.345 g$ and $62.439 g$ respectively.

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Ch. 15.1 - Prob. 15.1P Ch. 15.1 - Prob. 15.2P Ch. 15.2 - Prob. 15.3P Ch. 15.2 - Prob. 15.4P Ch. 15.3 - Prob. 15.5P Ch. 15.4 - Prob. 15.6P Ch. 15.5 - Prob. 15.7P Ch. 15.6 - Prob. 15.8P Ch. 15 - Prob. 1RQ Ch. 15 - Prob. 2RQ

Concept explainers

Videos

Explanation of Solution

Interpretation:

Mass of each ion found in $100 mL$ of $3.50 M K3AsO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

Mass of each ion found in $100 mL$ of $0.75 M NaHCO3$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

Mass of each ion found in $100 mL$ of $0.65 M (NH4)2SO4$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Want to see more full solutions like this?

Chapter 15 Solutions