Chapter 7, Problem 7.64E

Calculate the boiling and freezing points of water solutions that are $1.50\text{M}$ in the following solutes:

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

b. glycerol, a nonelectrolyte

c. ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{SO}}_4$, strong electrolyte

d. $\text{Al}{\left({\text{NO}}_3\right)}_{\text{3}}$, a strong electrolyte

Interpretation Introduction

(a)

Interpretation:

The boiling and freezing points of $1.50\text{M}$ water solution of $\text{KCl}$, 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 7.64E

The boiling and freezing points of $1.50\text{M}$ water solution of $\text{KCl}$, a strong electrolyte are $\text{101}\text{.6}°\text{C}$ and $-5.58°\text{C}$ 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 $\text{KCl}$ is a strong electrolyte it will completely dissociate in the solution and the value of $n$ for $\text{KCl}$ 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.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& \text{2}\times \text{0}\text{.52}°\text{C}/\text{M}\times 1.50\text{M}\\ & =& 1.56°\text{C}\end{array}$

The boiling point of water solution can be calculated by adding value the of $\Delta T_b$ to normal boiling point of water as shown below.

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+1.56°\text{C}\\ & =& 101\text{.56}°\text{C}\\ & \approx & \text{101}\text{.6}°\text{C}\end{array}$

Thus, the boiling point of water solution is $\text{101}\text{.6}°\text{C}$.

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 $\text{KCl}$ is a strong electrolyte it will completely dissociate in the solution and the value of $n$ for $\text{KCl}$ 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.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& \text{2}\times \text{1}\text{.86}°\text{C}/\text{M}\times 1.50\text{M}\\ & =& 5.58°\text{C}\end{array}$

The freezing point of water solution can be calculated by subtracting the value of $\Delta T_f$ from freezing point of water as shown below.

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-5.58°\text{C}\\ & =& -5.58°\text{C}\end{array}$

Thus, the freezing point of the water solution is $-5.58°\text{C}$.

Conclusion

The boiling and freezing points of $1.50\text{M}$ water solution of $\text{KCl}$, a strong electrolyte are $\text{101}\text{.6}°\text{C}$ and $-5.58°\text{C}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The boiling and freezing points of $1.50\text{M}$ water solution of glycerol, a nonelectrolyte is 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 7.64E

The boiling and freezing points of $1.50\text{M}$ water solution of glycerol, a nonelectrolyte are $\text{100}\text{.78}°\text{C}$ and $-2.79°\text{C}$ 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 glycerol 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.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 1\times \text{0}\text{.52}°\text{C}/\text{M}\times 1.50\text{M}\\ & =& 0.78°\text{C}\end{array}$

The boiling point of water solution can be calculated by adding value of $\Delta T_b$ to the normal boiling point of water as shown below.

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+0.78°\text{C}\\ & =& 100\text{.78}°\text{C}\end{array}$

Thus, the boiling point of water solution is $\text{100}\text{.78}°\text{C}$.

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 glyerol is a nonelectrolyte it will not dissociate in the solution and the value of $n$ for glycerol 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.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 1\times \text{1}\text{.86}°\text{C}/\text{M}\times 1.50\text{M}\\ & =& 2.79°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-2.79°\text{C}\\ & =& -2.79°\text{C}\end{array}$

Thus, the freezing point of water solution is $-2.79°\text{C}$.

Conclusion

The boiling and freezing points of $1.50\text{M}$ water solution of glycerol, a nonelectrolyte are $\text{100}\text{.78}°\text{C}$ and $-2.79°\text{C}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The boiling and freezing points of $1.50\text{M}$ water solution of ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{SO}}_4$, 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 7.64E

The boiling and freezing points of $1.50\text{M}$ water solution of ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{SO}}_4$, strong electrolyte are $\text{102}\text{.34}°\text{C}$ and $-8.37°\text{C}$ 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 ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{SO}}_4$ is a strong electrolyte it will dissociate in the solution completely and the value of $n$ for ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{SO}}_4$ 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.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 3\times \text{0}\text{.52}°\text{C}/\text{M}\times 1.50\text{M}\\ & =& 2.34°\text{C}\end{array}$

The boiling point of water solution can be calculated by adding value of $\Delta T_b$ to the normal boing point of water as shown below.

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+2.34°\text{C}\\ & =& 102\text{.34}°\text{C}\end{array}$

Thus, the boiling point of water solution is $\text{102}\text{.34}°\text{C}$.

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 ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{SO}}_4$ is a strong electrolyte it will dissociate in the solution completely and the value of $n$ for ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{SO}}_4$ 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.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 3\times \text{1}\text{.86}°\text{C}/\text{M}\times 1.50\text{M}\\ & =& 8.37°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-8.37°\text{C}\\ & =& -8.37°\text{C}\end{array}$

Thus, the freezing point of water solution is $-8.37°\text{C}$.

Conclusion

The boiling and freezing points of $1.50\text{M}$ water solution of ${\left({\text{NH}}_{\text{4}}\right)}_2{\text{SO}}_4$, strong electrolyte are $\text{102}\text{.34}°\text{C}$ and $-8.37°\text{C}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The boiling and freezing points of $1.50\text{M}$ water solution of $\text{Al}{\left({\text{NO}}_3\right)}_{\text{3}}$, 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 7.64E

The boiling and freezing points of $1.50\text{M}$ water solution of $\text{Al}{\left({\text{NO}}_3\right)}_{\text{3}}$, a strong electrolyte are $\text{103}\text{.12}°\text{C}$ and $-11.16°\text{C}$ 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 $\text{Al}{\left({\text{NO}}_3\right)}_{\text{3}}$ is a strong electrolyte it will dissociate in the solution completely and the value of $n$ for $\text{Al}{\left({\text{NO}}_3\right)}_{\text{3}}$ is $4$. 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.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 4\times \text{0}\text{.52}°\text{C}/\text{M}\times 1.50\text{M}\\ & =& 3.12°\text{C}\end{array}$

The boiling point of water solution can be calculated by adding value of $\Delta T_b$ to the normal boing point of water as shown below.

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+3.12°\text{C}\\ & =& 103\text{.12}°\text{C}\end{array}$

Thus, the boiling point of water solution is $\text{103}\text{.12}°\text{C}$.

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 $\text{Al}{\left({\text{NO}}_3\right)}_{\text{3}}$ is a strong electrolyte it will dissociate in the solution completely and the value of $n$ for $\text{Al}{\left({\text{NO}}_3\right)}_{\text{3}}$ is $4$. 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.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 4\times \text{1}\text{.86}°\text{C}/\text{M}\times 1.50\text{M}\\ & =& 11.16°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-11.16°\text{C}\\ & =& -11.16°\text{C}\end{array}$

Thus, the freezing point of water solution is $-11.16°\text{C}$.

Conclusion

The boiling and freezing points of $1.50\text{M}$ water solution of $\text{Al}{\left({\text{NO}}_3\right)}_{\text{3}}$, a strong electrolyte are $\text{103}\text{.12}°\text{C}$ and $-11.16°\text{C}$ respectively.