Chapter 7, Problem 7.65E

Calculate the boiling and freezing points of water solutions that are $1.15\text{M}$ in the following solutes:

a. $\text{KBr}$, a strong electrolyte

b. ethylene glycol, a nonelectrolyte

c. ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{CO}}_{\text{3}}$, strong electrolyte

d. ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$, a strong electrolyte

Interpretation Introduction

(a)

Interpretation:

The boiling and freezing points of $1.15\text{M}$ water solution of $\text{KBr}$, a strong electrolyte are to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.65E

The boiling and freezing points of $1.15\text{M}$ water solution of $\text{KBr}$, a strong electrolyte are $\text{101}\text{.2}°\text{C}$ and $-4.3°\text{C}$ respectively.

Explanation of Solution

The formula to calculate boiling point is given below as,

$\Delta T_b=n{K}_b\text{M}$

Where,

• $n$ is the number of ions in the solution.

• $K_b$ is the boiling point constant which is defined for a particular solvent.

• $\text{M}$ is the molarity of solution.

Since $\text{KBr}$ is a strong electrolyte it will completely dissociate in the solution and the value of $n$ for $\text{KBr}$ is $2$. The value of $K_b$ for water is $\text{0}\text{.52}°\text{C}/\text{M}$.

Substitute the value of $n$, $K_b$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& \text{2}\times \text{0}\text{.52}°\text{C}/\text{M}\times 1.15\text{M}\\ & =& 1.196°\text{C}\end{array}$

Now, the boiling point of water solution can be calculated by adding value of $\Delta T_b$ as given below.

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+1.196°\text{C}\\ & =& 101\text{.196}°\text{C}\\ & \approx & \text{101}\text{.2}°\text{C}\end{array}$

The formula to calculate freezing point is given below as,

$\Delta T_f=n{K}_f\text{M}$

Where,

• $n$ is the number of ions in the solution.

• $K_f$ is the freezing point constant which is defined for a particular solvent.

• $\text{M}$ is the molarity of solution.

Since $\text{KBr}$ is a strong electrolyte it will completely dissociate in the solution and the value of $n$ for $\text{KBr}$ is $2$. The value of $K_f$ for water is $\text{1}\text{.86}°\text{C}/\text{M}$.

Substitute the value of $n$, $K_f$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& \text{2}\times \text{1}\text{.86}°\text{C}/\text{M}\times 1.15\text{M}\\ & =& 4.278°\text{C}\end{array}$

Now, the freezing point of water solution can be calculated by subtracting value of $\Delta T_f$ as given below.

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-4.278°\text{C}\\ & =& -4.278°\text{C}\\ & \approx & -4.3°\text{C}\end{array}$

Conclusion

The boiling and freezing points of $1.15\text{M}$ water solution of $\text{KBr}$, a strong electrolyte are $\text{101}\text{.2}°\text{C}$ and $-4.3°\text{C}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The boiling and freezing points of $1.15\text{M}$ water solution of ethylene glycol, a nonelectrolyte, are to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.65E

The boiling and freezing points of $1.15\text{M}$ water solution of ethylene glycol, a nonelectrolyte are $\text{100}\text{.6}°\text{C}$ and $-2.1°\text{C}$ respectively.

Explanation of Solution

The formula to calculate boiling point is given below as,

$\Delta T_b=n{K}_b\text{M}$

Where,

• $n$ is the number of ions in the solution.

• $K_b$ is the boiling point constant which is defined for a particular solvent.

• $\text{M}$ is the molarity of solution.

Since ethylene glycol is a nonelectrolyte it will not dissociate in the solution and the value of $n$ for ethylene glycol is $1$. The value of $K_b$ for water is $\text{0}\text{.52}°\text{C}/\text{M}$.

Substitute the value of $n$, $K_b$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 1\times \text{0}\text{.52}°\text{C}/\text{M}\times 1.15\text{M}\\ & =& 0.598°\text{C}\end{array}$

Now, the boiling point of water solution can be calculated by adding value of $\Delta T_b$ as given below.

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+0.598°\text{C}\\ & =& 100\text{.598}°\text{C}\\ & \approx & \text{100}\text{.6}°\text{C}\end{array}$

The formula to calculate freezing point is given below as,

$\Delta T_f=n{K}_f\text{M}$

Where,

• $n$ is the number of ions in the solution.

• $K_f$ is the freezing point constant which is defined for a particular solvent.

• $\text{M}$ is the molarity of solution.

Since ethylene glycol is a nonelectrolyte it will not dissociate in the solution and the value of $n$ for ethylene glycol is $1$. The value of $K_f$ for water is $\text{1}\text{.86}°\text{C}/\text{M}$.

Substitute the value of $n$, $K_f$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 1\times \text{1}\text{.86}°\text{C}/\text{M}\times 1.15\text{M}\\ & =& 2.139°\text{C}\end{array}$

Now, the freezing point of water solution can be calculated by subtracting value of $\Delta T_f$ as given below.

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-2.139°\text{C}\\ & =& -2.139°\text{C}\\ & \approx & -2.1°\text{C}\end{array}$

Conclusion

The boiling and freezing points of $1.15\text{M}$ water solution of ethylene glycol, a nonelectrolyte are $\text{100}\text{.6}°\text{C}$ and $-2.1°\text{C}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The boiling and freezing points of $1.15\text{M}$ water solution of ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{CO}}_{\text{3}}$, a strong electrolyte, are to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.65E

The boiling and freezing points of $1.15\text{M}$ water solution of ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{CO}}_{\text{3}}$, strong electrolyte are $\text{101}\text{.8}°\text{C}$ and $-6.4°\text{C}$ respectively.

Explanation of Solution

The formula to calculate boiling point is given below as,

$\Delta T_b=n{K}_b\text{M}$

Where,

• $n$ is the number of ions in the solution.

• $K_b$ is the boiling point constant which is defined for a particular solvent.

• $\text{M}$ is the molarity of solution.

Since ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{CO}}_{\text{3}}$ is a strong electrolyte it will dissociate in the solution completely and the value of $n$ for ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{CO}}_{\text{3}}$ is $3$. The value of $K_b$ for water is $\text{0}\text{.52}°\text{C}/\text{M}$.

Substitute the value of $n$, $K_b$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 3\times \text{0}\text{.52}°\text{C}/\text{M}\times 1.15\text{M}\\ & =& 1.794°\text{C}\end{array}$

Now, the boiling point of water solution can be calculated by adding value of $\Delta T_b$ as given below.

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+1.794°\text{C}\\ & =& 101\text{.794}°\text{C}\\ & \approx & \text{101}\text{.8}°\text{C}\end{array}$

The formula to calculate freezing point is given below as,

$\Delta T_f=n{K}_f\text{M}$

Where,

• $n$ is the number of ions in the solution.

• $K_f$ is the freezing point constant which is defined for a particular solvent.

• $\text{M}$ is the molarity of solution.

Since ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{CO}}_{\text{3}}$ is a strong electrolyte it will dissociate in the solution completely and the value of $n$ for ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{CO}}_{\text{3}}$ is $3$. The value of $K_f$ for water is $\text{1}\text{.86}°\text{C}/\text{M}$.

Substitute the value of $n$, $K_f$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 3\times \text{1}\text{.86}°\text{C}/\text{M}\times 1.15\text{M}\\ & =& 6.417°\text{C}\end{array}$

Now, the freezing point of water solution can be calculated by subtracting value of $\Delta T_f$ as given below.

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-6.417°\text{C}\\ & =& -6.417°\text{C}\\ & \approx & -6.4°\text{C}\end{array}$

Conclusion

The boiling and freezing points of $1.15\text{M}$ water solution of ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{CO}}_{\text{3}}$, strong electrolyte are $\text{101}\text{.8}°\text{C}$ and $-6.4°\text{C}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The boiling and freezing points of $1.15\text{M}$ water solution of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$, a strong electrolyte are to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.65E

The boiling and freezing points of $1.15\text{M}$ water solution of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$, a strong electrolyte are $\text{103}°\text{C}$ and $-10.7°\text{C}$ respectively.

Explanation of Solution

The formula to calculate boiling point is given below as,

$\Delta T_b=n{K}_b\text{M}$

Where,

• $n$ is the number of ions in the solution.

• $K_b$ is the boiling point constant which is defined for a particular solvent.

• $\text{M}$ is the molarity of solution.

Since ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is a strong electrolyte it will dissociate in the solution completely and the value of $n$ for ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is $5$. The value of $K_b$ for water is $\text{0}\text{.52}°\text{C}/\text{M}$.

Substitute the value of $n$, $K_b$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 5\times \text{0}\text{.52}°\text{C}/\text{M}\times 1.15\text{M}\\ & =& 2.99°\text{C}\end{array}$

Now, the boiling point of water solution can be calculated by adding value of $\Delta T_b$ as given below.

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+2.99°\text{C}\\ & =& 102\text{.99}°\text{C}\\ & \approx & \text{103}°\text{C}\end{array}$

The formula to calculate freezing point is given below as,

$\Delta T_f=n{K}_f\text{M}$

Where,

• $n$ is the number of ions in the solution.

• $K_f$ is the freezing point constant which is defined for a particular solvent.

• $\text{M}$ is the molarity of solution.

Since ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is a strong electrolyte it will dissociate in the solution completely and the value of $n$ for ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is $5$. The value of $K_f$ for water is $\text{1}\text{.86}°\text{C}/\text{M}$.

Substitute the value of $n$, $K_f$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 5\times \text{1}\text{.86}°\text{C}/\text{M}\times 1.15\text{M}\\ & =& 10.695°\text{C}\end{array}$

Now, the freezing point of water solution can be calculated by subtracting value of $\Delta T_f$ as given below.

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-10.695°\text{C}\\ & =& -10.695°\text{C}\\ & \approx & -10.7°\text{C}\end{array}$

Conclusion

The boiling and freezing points of $1.15\text{M}$ water solution of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$, a strong electrolyte are $\text{103}°\text{C}$ and $-10.7°\text{C}$ respectively.