Chapter 4, Problem 4.28P

(a)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $37.00\text{ mL}$ of $0.250M$ potassium chloride to $150.00\text{ mL}$ 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.

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution prepared by diluting $25.71\text{ mL}$ of $0.0706M$ ammonium sulfate to $500.00\text{ mL}$ 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.

(c)

Interpretation Introduction

Interpretation:

Molarity of sodium ion in a solution made by mixing $3.58\text{ mL}$ of $0.348M$ sodium chloride with $500\text{ mL}$ of $6.81\times {10}^{-2}M$ sodium sulfate is to be calculated.

Concept introduction:

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

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

$\text{Moles of ion from compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{volume of solution}\left(\text{L}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ \left(\frac{\text{molarity}\text{of}\text{solution}\left(\text{mol}\right)}{\text{1L of solution}}\right)\\ \left(\frac{\text{mole}\text{of}\text{charge on ion}}{1\text{mol of compound}}\right)\end{array}\right]$ (3)