Chapter 24, Problem 119A

(a)

Interpretation Introduction

Interpretation: To calculate the boiling point elevation for the given aqueous solution.

Concept Introduction: The magnitude of the boiling point elevation is in direct relation to the molal concentration. It can be expressed as follows:

  $\Delta T_b\propto m$

This can also be expressed as follows:

  $\Delta T_b=K_b\times m$   .....(1)

Where $\Delta T_b$ represents the boiling point elevation, $K_b$ represents the molal boiling point elevation constant, and $m$ represents the molal concentration.

Answer to Problem 119A

The boiling point elevation for $0.507m\text{ NaCl}$ solution is $\text{0}\text{.519}°\text{C}$ .

Explanation of Solution

The given aqueous solution is $0.507m\text{ NaCl}$ .

Each formula unit of $\text{NaCl}$ dissociates into two particles, thus the molal concentration changes to $2\times 0.507m=1.014m$ .

The molal boiling point elevation for water is $0.512°\text{C}/m$ .

Substitute $0.512°\text{C}/m$ for $K_b$ , $1.014m$ for $m$ in equation (1) in order to calculate the boiling point elevation.

  $\begin{array}{c}\Delta T_b=K_b\times m\\ =\frac{0.512°\text{C}}{\cancel{m}}\times 1.014\cancel{m}\\ =0.519°\text{C}\end{array}$

Thus, the boiling point elevation for $0.507m\text{ NaCl}$ solution is $\text{0}\text{.519}°\text{C}$ .

(b)

Interpretation Introduction

Interpretation: To calculate the boiling point elevation for the given aqueous solution.

Concept Introduction: The magnitude of the boiling point elevation is in direct relation to the molal concentration. It can be expressed as follows:

  $\Delta T_b\propto m$

This can also be expressed as follows:

  $\Delta T_b=K_b\times m$   .....(1)

Where $\Delta T_b$ represents the boiling point elevation, $K_b$ represents the molal boiling point elevation constant, and $m$ represents the molal concentration.

Answer to Problem 119A

The boiling point elevation for $0.204m{\text{ NH}}_4\text{Cl}$ solution is $\text{0}\text{.209}°\text{C}$ .

Explanation of Solution

The given aqueous solution is $0.204m{\text{ NH}}_4\text{Cl}$ .

Each formula unit of ${\text{NH}}_4\text{Cl}$ dissociates into two particles, thus the molal concentration changes to $2\times 0.204m=0.408m$ .

The molal boiling point elevation for water is $0.512°\text{C}/m$ .

Substitute $0.512°\text{C}/m$ for $K_b$ , $0.408m$ for $m$ in equation (1) in order to calculate the boiling point elevation.

  $\begin{array}{c}\Delta T_b=K_b\times m\\ =\frac{0.512°\text{C}}{\cancel{m}}\times 0.408\cancel{m}\\ =0.209°\text{C}\end{array}$

Thus, the boiling point elevation for $0.204m{\text{ NH}}_4\text{Cl}$ solution is $\text{0}\text{.209}°\text{C}$ .

(c)

Interpretation Introduction

Interpretation: To calculate the boiling point elevation for the given aqueous solution.

Concept Introduction: The magnitude of the boiling point elevation is in direct relation to the molal concentration. It can be expressed as follows:

  $\Delta T_b\propto m$

This can also be expressed as follows:

  $\Delta T_b=K_b\times m$   .....(1)

Where $\Delta T_b$ represents the boiling point elevation, $K_b$ represents the molal boiling point elevation constant and $m$ represents the molal concentration.

Answer to Problem 119A

The boiling point elevation for $0.155m{\text{ CaCl}}_2$ solution is $\text{0}\text{.238}°\text{C}$ .

Explanation of Solution

The given aqueous solution is $0.155m{\text{ CaCl}}_2$ .

Each formula unit of ${\text{CaCl}}_2$ dissociates into three particles, thus the molal concentration changes to $3\times 0.155m=0.465m$ .

The molal boiling point elevation for water is $0.512°\text{C}/m$ .

Substitute $0.512°\text{C}/m$ for $K_b$ , $0.465m$ for $m$ in equation (1) in order to calculate the boiling point elevation.

  $\begin{array}{c}\Delta T_b=K_b\times m\\ =\frac{0.512°\text{C}}{\cancel{m}}\times 0.465\cancel{m}\\ =0.238°\text{C}\end{array}$

Thus, the boiling point elevation for $0.155m{\text{ CaCl}}_2$ solution is $\text{0}\text{.238}°\text{C}$ .

(d)

Interpretation Introduction

Interpretation: To calculate the boiling point elevation for the given aqueous solution.

Concept Introduction: The magnitude of the boiling point elevation is in direct relation to the molal concentration. It can be expressed as follows:

  $\Delta T_b\propto m$

This can also be expressed as follows:

  $\Delta T_b=K_b\times m$   .....(1)

Where $\Delta T_b$ represents the boiling point elevation, $K_b$ represents the molal boiling point elevation constant and $m$ represents the molal concentration.

Answer to Problem 119A

The boiling point elevation for $0.222m{\text{ NaHSO}}_{\text{4}}$ solution is $\text{0}\text{.227}°\text{C}$ .

Explanation of Solution

The given aqueous solution is $0.222m{\text{ NaHSO}}_{\text{4}}$ .

Each formula unit of ${\text{NaHSO}}_4$ dissociates into two particles, thus the molal concentration changes to $2\times 0.222m=0.444m$ .

The molal boiling point elevation for water is $0.512°\text{C}/m$ .

Substitute $0.512°\text{C}/m$ for $K_b$ , $0.444m$ for $m$ in equation (1) in order to calculate the boiling point elevation.

  $\begin{array}{c}\Delta T_b=K_b\times m\\ =\frac{0.512°\text{C}}{\cancel{m}}\times 0.444\cancel{m}\\ =0.227°\text{C}\end{array}$

Thus, the boiling point elevation for $0.222m{\text{ NaHSO}}_{\text{4}}$ solution is $\text{0}\text{.227}°\text{C}$ .