Chapter 7, Problem 7.67E

Interpretation Introduction

(a)

Interpretation:

The boiling and freezing points of a solution containing $50.0\text{g}$ of ${\text{H}}_2{\text{SO}}_4$, a strong electrolyte (both $\text{H}$ s dissociate), per $250.\text{mL}$ 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 7.67E

The boiling and freezing points of a solution containing $50.0\text{g}$ of ${\text{H}}_2{\text{SO}}_4$, a strong electrolyte (both $\text{H}$ s dissociate), per $250.\text{mL}$ are $\text{103}\text{.18}°\text{C}$ and $-11.38°\text{C}$ respectively.

Explanation of Solution

The formula to calculate number of moles of solutes is given below as,

$\text{Moles}\text{of}\text{solute}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$

The molar mass of ${\text{H}}_2{\text{SO}}_4$ can be calculated as follows.

$\begin{array}{rll}{\text{H}}_2{\text{SO}}_4& =& \left(1\times \text{S}\right)+\left(2\times \text{H}\right)+\left(4\times \text{O}\right)\\ & =& \left(1\times 32\right)+\left(2\times 1\right)+\left(4\times 16\right)\\ & =& 98\text{g}/\text{mol}\end{array}$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Moles}\text{of}{\text{H}}_2{\text{SO}}_4& =& \frac{\text{Given}\text{mass}\text{of}{\text{H}}_2{\text{SO}}_4}{\text{Molar}\text{mass}\text{of}{\text{H}}_2{\text{SO}}_4}\\ & =& \frac{50\text{g}}{98\text{g}/\text{mol}}\\ & =& 0.51\text{mol}\end{array}$

The formula to calculate molarity is given below as,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}}{\text{liters}\text{of}\text{solution}}$

Substitute the value of number of moles of solute and volume of solution in the above equation as follows.

$\begin{array}{rll}\text{M}& =& \frac{\text{moles}\text{of}{\text{H}}_2{\text{SO}}_4}{\text{liters}\text{of}{\text{H}}_2{\text{SO}}_4\text{solution}}\\ & =& \frac{0.51\text{mol}}{250\text{mL}\times \frac{1\text{L}}{1000\text{mL}}}\\ & =& 2.04\text{M}\end{array}$

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 ${\text{H}}_2{\text{SO}}_4$ is a strong electrolyte, it will dissociate in the solution completely and the value of $n$ for ${\text{H}}_2{\text{SO}}_4$ 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.

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 3\times \text{0}\text{.52}°\text{C}/\text{M}\times 2.04\text{M}\\ & =& 3.18°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+3.18°\text{C}\\ & =& 103\text{.18}°\text{C}\end{array}$

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 ${\text{H}}_2{\text{SO}}_4$ is a strong electrolyte, it will dissociate in the solution completely and the value of $n$ for urea 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.

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 3\times \text{1}\text{.86}°\text{C}/\text{M}\times 2.04\text{M}\\ & =& 11.38°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-11.38°\text{C}\\ & =& -11.38°\text{C}\end{array}$

Conclusion

The boiling and freezing points of a solution containing $50.0\text{g}$ of ${\text{H}}_2{\text{SO}}_4$, a strong electrolyte (both $\text{H}$ s dissociate), per $250.\text{mL}$ are $\text{103}\text{.18}°\text{C}$ and $-11.38°\text{C}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The boiling and freezing points of a solution containing $200.\text{g}$ of table sugar $\left({\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\right)$, a nonelectrolyte, per $250.\text{mL}$ 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 7.67E

The boiling and freezing points of a solution containing $200.\text{g}$ of table sugar $\left({\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\right)$, a nonelectrolyte, per $250.\text{mL}$ are $\text{101}\text{.21}°\text{C}$ and $-4.33°\text{C}$ respectively.

Explanation of Solution

The formula to calculate number of moles of solutes is given below as,

$\text{Moles}\text{of}\text{solute}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$

The molar mass of sugar can be calculated as follows.

$\begin{array}{rll}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}& =& \left(12\times \text{C}\right)+\left(22\times \text{H}\right)+\left(11\times \text{O}\right)\\ & =& \left(12\times 12\right)+\left(22\times 1\right)+\left(11\times 16\right)\\ & =& 342\text{g}/\text{mol}\end{array}$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Moles}\text{of}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}& =& \frac{\text{Given}\text{mass}\text{of}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}\\ & =& \frac{200\text{g}}{342\text{g}/\text{mol}}\\ & =& 0.5847\text{mol}\end{array}$

The formula to calculate molarity is given below as,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}}{\text{liters}\text{of}\text{solution}}$

Substitute the value of number of moles of solute and volume of solution in the above equation as follows.

$\begin{array}{rll}\text{M}& =& \frac{\text{moles}\text{of}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{12}}}{\text{liters}\text{of}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{12}}\text{solution}}\\ & =& \frac{0.5847\text{mol}}{250\text{mL}\times \frac{1\text{L}}{1000\text{mL}}}\\ & =& 2.33\text{M}\end{array}$

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

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 1\times \text{0}\text{.52}°\text{C}/\text{M}\times 2.33\text{M}\\ & =& 1.21°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Boiling point}& =& 100°\text{C}+1.21°\text{C}\\ & =& 101\text{.21}°\text{C}\end{array}$

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

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 1\times \text{1}\text{.86}°\text{C}/\text{M}\times 2.33\text{M}\\ & =& 4.33°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Freezing point}& =& 0°\text{C}-4.33°\text{C}\\ & =& -4.33°\text{C}\end{array}$

Conclusion

The boiling and freezing points of a solution containing $200.\text{g}$ of table sugar $\left({\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\right)$, a nonelectrolyte, per $250.\text{mL}$ are $\text{101}\text{.21}°\text{C}$ and $-4.33°\text{C}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The boiling and freezing points of a solution containing $75.0\text{g}$ of octanoic acid $\left({\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}\right)$, a nonelectrolyte, in enough benzene to give $250.\text{mL}$ of solution 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 7.67E

The boiling and freezing points of a solution containing $75.0\text{g}$ of octanoic acid $\left({\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}\right)$, a nonelectrolyte, in enough benzene to give $250.\text{mL}$ of solution are $\text{85}\text{.36}°\text{C}$ and $-4.69°\text{C}$ respectively.

Explanation of Solution

The formula to calculate number of moles of solutes is given below as,

$\text{Moles}\text{of}\text{solute}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$

The molar mass of ethylene glycol can be calculated as follows.

$\begin{array}{rll}{\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}& =& \left(8\times \text{C}\right)+\left(16\times \text{H}\right)+\left(2\times \text{O}\right)\\ & =& \left(8\times 12\right)+\left(16\times 1\right)+\left(2\times 16\right)\\ & =& 144\text{g}/\text{mol}\end{array}$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Moles}\text{of}{\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}& =& \frac{\text{Given}\text{mass}\text{of}{\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}}{\text{Molar}\text{mass}\text{of}{\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}}\\ & =& \frac{75\text{g}}{144\text{g}/\text{mol}}\\ & =& 0.5208\text{mol}\end{array}$

The formula to calculate molarity is given below as,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}}{\text{liters}\text{of}\text{solution}}$

Substitute the value of number of moles of solute and volume of solution in the above equation as follows.

$\begin{array}{rll}\text{M}& =& \frac{\text{moles}\text{of}{\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}}{\text{liters}\text{of}{\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}\text{solution}}\\ & =& \frac{0.5208\text{mol}}{250\text{mL}\times \frac{1\text{L}}{1000\text{mL}}}\\ & =& 2.08\text{M}\end{array}$

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 octanoic acid is a nonelectrolyte, it will not dissociate in the solution and the value of $n$ for octanoic acid is $1$. The value of $K_b$ for benzene is $\text{2}\text{.53}°\text{C}/\text{M}$.

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

$\begin{array}{rll}\Delta T_b& =& n{K}_b\text{M}\\ & =& 1\times \text{2}\text{.53}°\text{C}/\text{M}\times 2.08\text{M}\\ & =& 5.26°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Boiling point}& =& 80.1°\text{C}+5.26°\text{C}\\ & =& \text{85}\text{.36}°\text{C}\end{array}$

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 octanoic acid is a nonelectrolyte, it will not dissociate in the solution and the value of $n$ for octanoic acid is $1$. The value of $K_f$ for benzene is $\text{4}\text{.90}°\text{C}/\text{M}$.

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

$\begin{array}{rll}\Delta T_f& =& n{K}_f\text{M}\\ & =& 1\times \text{4}\text{.90}°\text{C}/\text{M}\times 2.08\text{M}\\ & =& 10.19°\text{C}\end{array}$

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

$\begin{array}{rll}\text{Freezing point}& =& 5.5°\text{C}-10.19°\text{C}\\ & =& -4.69°\text{C}\end{array}$

Conclusion

The boiling and freezing points of a solution containing $75.0\text{g}$ of octanoic acid $\left({\text{C}}_8{\text{H}}_{\text{16}}{\text{O}}_{\text{2}}\right)$, a nonelectrolyte, in enough benzene to give $250.\text{mL}$ of solution are $\text{85}\text{.36}°\text{C}$ and $-4.69°\text{C}$ respectively.