Chapter 15, Problem 27A

(a)

Interpretation Introduction

Interpretation:

The number of moles and the number of grams of the given solute in a given solutionhave to be calculated.

Concept Introduction:

The composition of a solution can be defined by expressing its concentration. The concentrations of solutions can be expressed in different ways, which are involved in the quantity of solute and the quantity of solution or solvent. Various methodsare usedto describe the concentration of the solution quantitatively. Some commonly used quantitative concentration terms are percent by mass, percent by volume, molarity, molality and mole fraction.

Molarity: Molarity is defined as the number of moles of solute present in one liter of the solution.

In expression, $\text{Molarity}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Again $\text{Number}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}$

Answer to Problem 27A

	Number of moles	Number of grams
127 mL of 0.105 M HNO3	0.0133 mol	0.84 gm

Explanation of Solution

$\text{Molarity}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Data given: Molarity of HNO₃ solution= 0.105 M

      Volume of HNO₃ solution = 127 mL = 0.127 L

 $\begin{array}{c}\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{H}\text{N}{\text{O}}_{\text{3}}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}\end{array}$

  $\begin{array}{c}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}=\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}\times \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{H}\text{N}{\text{O}}_{\text{3}}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})\end{array}$

  $\begin{array}{c}=\text{0}.\text{1}\text{0}\text{5}\text{M}\times \text{0}.\text{1}\text{2}\text{7}\text{L}\end{array}$$\begin{array}{c}=\text{0}.\text{0}\text{1}\text{3}\text{3}\text{m}\text{o}\text{l}\end{array}$

$\begin{array}{c}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}}\end{array}$

      $\begin{array}{c}\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}=\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}\times \text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{H}\text{N}{\text{O}}_{\text{3}}\end{array}$

      $\begin{array}{c}=\text{0}.\text{0}\text{1}\text{3}\text{3}\text{m}\text{o}\text{l}\times \text{6}\text{3}.\text{0}\text{1}\text{g}\text{m}/\text{m}\text{o}\text{l}\end{array}$$\begin{array}{c}=\text{0}.\text{8}\text{4}\text{g}\text{m}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The number of moles and the number of grams of the given solute in a given solutionhave to be calculated.

Concept Introduction:

The composition of a solution can be defined by expressing its concentration. The concentrations of solutions can be expressed in different ways, which are involved in the quantity of solute and the quantity of solution or solvent. Various methods are used to describe the concentration of the solution quantitatively. Some commonly used quantitative concentration terms are percent by mass, percent by volume, molarity, molality and mole fraction.

Molarity: Molarity is defined as the number of moles of solute present in one liter of the solution.

In expression, $\text{Molarity}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Again $\text{Number}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}$

Answer to Problem 27A

	Number of moles	Number of grams
155 mL of 15.1 M NH3	2.34 mol	39.79 gm

Explanation of Solution

$\text{Molarity}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Data given: Molarity of NH₃ solution= 15.1 M

      Volume of NH₃ solution = 155 mL = 0.155 L

  $\begin{array}{c}\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{N}{\text{H}}_{\text{3}}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}\end{array}$

  $\begin{array}{c}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}=\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}\times \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{N}{\text{H}}_{\text{3}}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})\end{array}$

  $\begin{array}{c}=\text{1}\text{5}.\text{1}\text{M}\times \text{0}.\text{1}\text{5}\text{5}\text{L}\end{array}$$\begin{array}{c}=\text{2}.\text{3}\text{4}\text{m}\text{o}\text{l}\end{array}$

$\begin{array}{c}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}}\end{array}$

    $\begin{array}{c}\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}=\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}\times \text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{N}{\text{H}}_{\text{3}}\end{array}$

    $\begin{array}{c}=\text{2}.\text{3}\text{4}\text{m}\text{o}\text{l}\times \text{1}\text{7}\text{g}\text{m}/\text{m}\text{o}\text{l}\end{array}$$\begin{array}{c}=\text{3}\text{9}.\text{7}\text{9}\text{g}\text{m}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The number of moles and the number of grams of the given solute in a given solutionhave to be calculated.

Concept Introduction:

The composition of a solution can be defined by expressing its concentration. The concentrations of solutions can be expressed in different ways, which are involved in the quantity of solute and the quantity of solution or solvent. Various methods are used to describe the concentration of the solution quantitatively. Some commonly used quantitative concentration terms are percent by mass, percent by volume, molarity, molality and mole fraction.

Molarity: Molarity is defined as the number of moles of solute present in one liter of the solution.

In expression, $\text{Molarity}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Again $\text{Number}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}$

Answer to Problem 27A

	Number of moles	Number of grams
2.51 L of 2.01 × 10-3 M KSCN	0.005 mol	0.49 gm

Explanation of Solution

  $\text{Molarity}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Data given: Molarity of KSCN solution= 2.01 × 10-³M

      The volume of KSCN solution = 2.51 L

$\begin{array}{c}\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{K}\text{S}\text{C}\text{N}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}\end{array}$

$\begin{array}{c}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}=\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}\times \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{K}\text{S}\text{C}\text{N}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})\end{array}$

  $\begin{array}{c}=\text{2}.\text{0}\text{1}\times \text{1}{\text{0}}^{-\text{3}}\text{M}\times \text{2}.\text{5}\text{1}\text{L}\end{array}$ $\begin{array}{c}=\text{0}.\text{0}\text{0}\text{5}\text{m}\text{o}\text{l}\end{array}$

$\begin{array}{c}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}}\end{array}$

      $\begin{array}{c}\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}=\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}\times \text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{K}\text{S}\text{C}\text{N}\end{array}$

      $\begin{array}{c}=\text{0}.\text{0}\text{0}\text{5}\text{m}\text{o}\text{l}\times \text{9}\text{7}.\text{1}\text{8}\text{g}\text{m}/\text{m}\text{o}\text{l}\end{array}$ $\begin{array}{c}=\text{0}.\text{4}\text{9}\text{g}\text{m}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The number of moles and the number of grams of the given solute in a given solutionhave to be calculated.

Concept Introduction:

The composition of a solution can be defined by expressing its concentration. The concentrations of solutions can be expressed in different ways, which are involved in the quantity of solute and the quantity of solution or solvent. Various methods are used to describe the concentration of the solution quantitatively. Some commonly used quantitative concentration terms are percent by mass, percent by volume, molarity, molality and mole fraction.

Molarity: Molarity is defined as the number of moles of solute present in one liter of the solution.

In expression, $\text{Molarity}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Again $\text{Number}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}$

Answer to Problem 27A

	Number of moles	Number of grams
12.2 mL of 2.45 M HCl	0.03 mol	1.1 gm

Explanation of Solution

  $\text{Molarity}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{t}\text{h}\text{e} \text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Data given: Molarity of HCl solution= 2.45 M

      Volume of HCl solution= 12.2 mL = 0.0122 L

  $\begin{array}{c}\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{H}\text{C}\text{l}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{H}\text{C}\text{l}}{ \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{H}\text{C}\text{l}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}\end{array}$

  $\begin{array}{c}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{H}\text{C}\text{l}=\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{H}\text{C}\text{l}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}\times \text{V}\text{o}\text{l}\text{u}\text{m}\text{e} \text{o}\text{f} \text{H}\text{C}\text{l}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})\end{array}$

  $\begin{array}{c}=\text{2}.\text{4}\text{5}\text{M}\times \text{0}.\text{0}\text{1}\text{2}\text{2}\text{L}\end{array}$$\begin{array}{c}=\text{0}.\text{0}\text{3}\text{m}\text{o}\text{l}\end{array}$

  $\begin{array}{c}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{H}\text{C}\text{l}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{H}\text{C}\text{l}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{H}\text{C}\text{l}}\end{array}$

      $\begin{array}{c}\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{H}\text{C}\text{l}=\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{H}\text{C}\text{l}\times \text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{H}\text{C}\text{l}\end{array}$

      $\begin{array}{c}=\text{0}.\text{0}\text{3}\text{m}\text{o}\text{l}\times \text{3}\text{6}.\text{5}\text{g}\text{m}/\text{m}\text{o}\text{l}\end{array}$$\begin{array}{c}=\text{1}.\text{1}\text{g}\text{m}\end{array}$