Molarity of a solution prepared by diluting 37.00 mL of 0.250 M potassium chloride to 150.00 mL is to be calculated. Concept introduction: Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed. The expression to relate the molarity of a concentrated and dilute solution is: M 1 V 1 = M 2 V 2 (1) Here, M 1 is the molarity of the dilute solution. V 1 is the volume of dilute solution. M 2 is the molarity of the concentrated solution. V 2 is the volume of the concentrated solution.

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Answer 1

Question

Chapter 4, Problem 4.28P

(a)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $37.00 mL$ of $0.250 M$ potassium chloride to $150.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $25.71 mL$ of $0.0706 M$ ammonium sulfate to $500.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(c)

Interpretation Introduction

Interpretation:

Molarity of sodium ion in a solution made by mixing $3.58 mL$ of $0.348 M$ sodium chloride with $500 mL$ of $6.81×10−2 M$ sodium sulfate is to be calculated.

Concept introduction:

Molarity $(M)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $mol/L$ .

The expression to calculate the moles of ions from compound when molarity of solution and volume of solution are given is as follows:

$Moles of ion from compound(mol)=[volume of solution(L)(1000 mL1 L)(molarity of solution(mol)1L of solution)(mole of charge on ion1 mol of compound)]$ (3)

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Answer 2

Question

Chapter 4, Problem 4.28P

(a)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $37.00 mL$ of $0.250 M$ potassium chloride to $150.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $25.71 mL$ of $0.0706 M$ ammonium sulfate to $500.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(c)

Interpretation Introduction

Interpretation:

Molarity of sodium ion in a solution made by mixing $3.58 mL$ of $0.348 M$ sodium chloride with $500 mL$ of $6.81×10−2 M$ sodium sulfate is to be calculated.

Concept introduction:

Molarity $(M)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $mol/L$ .

The expression to calculate the moles of ions from compound when molarity of solution and volume of solution are given is as follows:

$Moles of ion from compound(mol)=[volume of solution(L)(1000 mL1 L)(molarity of solution(mol)1L of solution)(mole of charge on ion1 mol of compound)]$ (3)

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Answer 3

Question

Chapter 4, Problem 4.28P

(a)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $37.00 mL$ of $0.250 M$ potassium chloride to $150.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $25.71 mL$ of $0.0706 M$ ammonium sulfate to $500.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(c)

Interpretation Introduction

Interpretation:

Molarity of sodium ion in a solution made by mixing $3.58 mL$ of $0.348 M$ sodium chloride with $500 mL$ of $6.81×10−2 M$ sodium sulfate is to be calculated.

Concept introduction:

Molarity $(M)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $mol/L$ .

The expression to calculate the moles of ions from compound when molarity of solution and volume of solution are given is as follows:

$Moles of ion from compound(mol)=[volume of solution(L)(1000 mL1 L)(molarity of solution(mol)1L of solution)(mole of charge on ion1 mol of compound)]$ (3)

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Answer 4

Question

Chapter 4, Problem 4.28P

(a)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $37.00 mL$ of $0.250 M$ potassium chloride to $150.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $25.71 mL$ of $0.0706 M$ ammonium sulfate to $500.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(c)

Interpretation Introduction

Interpretation:

Molarity of sodium ion in a solution made by mixing $3.58 mL$ of $0.348 M$ sodium chloride with $500 mL$ of $6.81×10−2 M$ sodium sulfate is to be calculated.

Concept introduction:

Molarity $(M)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $mol/L$ .

The expression to calculate the moles of ions from compound when molarity of solution and volume of solution are given is as follows:

$Moles of ion from compound(mol)=[volume of solution(L)(1000 mL1 L)(molarity of solution(mol)1L of solution)(mole of charge on ion1 mol of compound)]$ (3)

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Answer 5

Chapter 4, Problem 4.28P

(a)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $37.00 mL$ of $0.250 M$ potassium chloride to $150.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $25.71 mL$ of $0.0706 M$ ammonium sulfate to $500.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

(c)

Interpretation Introduction

Interpretation:

Molarity of sodium ion in a solution made by mixing $3.58 mL$ of $0.348 M$ sodium chloride with $500 mL$ of $6.81×10−2 M$ sodium sulfate is to be calculated.

Concept introduction:

Molarity $(M)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $mol/L$ .

The expression to calculate the moles of ions from compound when molarity of solution and volume of solution are given is as follows:

$Moles of ion from compound(mol)=[volume of solution(L)(1000 mL1 L)(molarity of solution(mol)1L of solution)(mole of charge on ion1 mol of compound)]$ (3)

Answer 6

Chapter 4, Problem 4.28P

(a)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $37.00 mL$ of $0.250 M$ potassium chloride to $150.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

Answer 7

Chapter 4, Problem 4.28P

(a)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $37.00 mL$ of $0.250 M$ potassium chloride to $150.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

Answer 8

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $25.71 mL$ of $0.0706 M$ ammonium sulfate to $500.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

Answer 9

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $25.71 mL$ of $0.0706 M$ ammonium sulfate to $500.00 mL$ is to be calculated.

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.

The expression to relate the molarity of a concentrated and dilute solution is:

$M1 V1=M2 V2$ (1)

Here,

$M1$ is the molarity of the dilute solution.

$V1$ is the volume of dilute solution.

$M2$ is the molarity of the concentrated solution.

$V2$ is the volume of the concentrated solution.

Answer 10

(c)

Interpretation Introduction

Interpretation:

Molarity of sodium ion in a solution made by mixing $3.58 mL$ of $0.348 M$ sodium chloride with $500 mL$ of $6.81×10−2 M$ sodium sulfate is to be calculated.

Concept introduction:

Molarity $(M)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $mol/L$ .

The expression to calculate the moles of ions from compound when molarity of solution and volume of solution are given is as follows:

$Moles of ion from compound(mol)=[volume of solution(L)(1000 mL1 L)(molarity of solution(mol)1L of solution)(mole of charge on ion1 mol of compound)]$ (3)

Answer 11

(c)

Interpretation Introduction

Interpretation:

Molarity of sodium ion in a solution made by mixing $3.58 mL$ of $0.348 M$ sodium chloride with $500 mL$ of $6.81×10−2 M$ sodium sulfate is to be calculated.

Concept introduction:

Molarity $(M)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $mol/L$ .

The expression to calculate the moles of ions from compound when molarity of solution and volume of solution are given is as follows:

$Moles of ion from compound(mol)=[volume of solution(L)(1000 mL1 L)(molarity of solution(mol)1L of solution)(mole of charge on ion1 mol of compound)]$ (3)

Answer 12

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Answer 13

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Answer 14

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Answer 15

Ch. 4.1 - (a) Which strong electrolyte is dissolved in water...Ch. 4.1 - (a) Make a drawing similar to that in Follow-up...Ch. 4.1 - What amount (mol) of each ion is in each...Ch. 4.1 - Prob. 4.2BFP Ch. 4.1 - Prob. 4.3AFP Ch. 4.1 - Prob. 4.3BFP Ch. 4.1 - Prob. 4.4AFP Ch. 4.1 - Prob. 4.4BFP Ch. 4.1 - What amount (mol) of each ion is in 1.32 L of 0.55...Ch. 4.1 - What is the molarity of aluminum ion in a solution...

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