Chapter 11, Problem 92E

Interpretation Introduction

Interpretation: The freezing point and the boiling point of the given solutions are should be calculated.

Concept introduction:

• Elevation of boiling point:

  The boiling point of the solution is increases when the solute is dissolved in the solvent is called Elevation of boiling point. it is one of the colligative Properties thus,

                                           $\begin{array}{l}\text{ΔT}\text{=}{\text{iK}}_{\text{b}}{\text{m}}_{\text{solute}}\mathrm{......}(1)\\ \text{ΔT is boiling-point elevation}\\ {\text{K}}_{\text{b}}\text{is}\text{molal boiling-point elevation constant}\\ \text{m}\text{is }\text{molality of the solute}\\ \text{i}\text{is}\text{the}\text{van't Hoff factor}\end{array}$

• Depression in freezing point:

  The freezing point the solution is decreases when the solute is dissolved in the solvent is called Elevation of boiling point. it is one of the colligative Properties thus

                                        $\begin{array}{l}\text{ΔT}\text{=}{\text{iK}}_{\text{f}}{\text{m}}_{\text{solute}}\mathrm{......}(2)\\ \text{ΔT is boiling-point elevation}\\ {\text{K}}_{\text{f}}\text{is}\text{molal freezing-point depression constant}\\ \text{m}\text{is }\text{molality of the solute}\\ \text{i}\text{is}\text{the}\text{van't Hoff factor}\end{array}$

Answer to Problem 92E

• The boiling point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ is $\text{100}\text{.069°C}$
• The freezing point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ is $\text{-0}\text{.25°C}$
• The boiling point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ is $\text{100}\text{.087°C}$
• The freezing point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ is $\text{-0}\text{.32°C}$

Explanation of Solution

Record the given data,

           Molality of ${\text{MgCl}}_{\text{2}}\text{= 0}\text{.050}\text{m}$

           Molality of ${\text{FeCl}}_{\text{3}}=\text{0}\text{.050}\text{m}$

To calculate the boiling point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ solution.

$\text{Molal boiling-point elevation constant}$ of water is $\text{0}\text{.51 °C/molal}$.

                          $\begin{array}{l}{\text{van't Hoff factor of MgCl}}_{\text{2}}\text{is}\text{2}\text{.7}\\ \text{ΔT}\text{=}2.7\text{×0}\text{.51 °C/molal×}\text{0}\text{.050}\text{molal}\\ \text{=}0.069\text{°C}\\ \text{=}\text{100°C+0}\text{.069°C}\\ {\text{T}}_{\text{b}}\text{=}\text{100}\text{.069°C}\\ \end{array}$

• The given values are plugging in to equation 1 to give elevation of boiling point this was added to boiling point of (solvent) water $\text{(100°C)}$ to give the boiling point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ solution.
• The boiling point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ solution is $\text{100}\text{.069°C}$

To calculate the freezing point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ solution

$\text{Molal freezing-point depression constant}\text{of water is 1}\text{.86}°\text{C/molal}$

                                  $\begin{array}{l}{\text{van't Hoff factor of MgCl}}_{\text{2}}\text{is}\text{2}\text{.7}\\ \text{ΔT=}2.7\text{×1}\text{.86 °C/molal×}\text{0}\text{.050}\text{molal}\\ \text{=}\text{0}\text{.25°C}\\ {\text{T}}_{\text{f}}\text{=}\text{0}\text{.00°C-0}\text{.25°C}\\ \text{=}\text{-25}\text{.0°C}\end{array}$

• The given values are plugging in to equation 1 to give depression in freezing point this was subtract from freezing point of (solvent) water $\text{(100°C)}$ to give the boiling point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ solution.
• The freezing point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ solution is $-25.\text{0°C}$ .

To calculate the boiling point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ solution.

$\text{Molal boiling-point elevation constant}$ of water is $\text{0}\text{.51 °C/molal}$

                              $\begin{array}{l}{\text{van't Hoff factor of FeCl}}_3\text{is}3.\text{4}\\ \text{ΔT}\text{=}3.4\text{×0}\text{.51 °C/molal×}\text{0}\text{.050}\text{molal}\\ \text{=}\text{0}\text{.087°C}\\ \text{=}\text{100°C+0}\text{.087°C}\\ {\text{T}}_{\text{b}}\text{=}\text{100}\text{.087°C}\end{array}$

• The given values are plugging in to equation 1 to give elevation of boiling point this was added to boiling point of (solvent) water $\text{(100°C)}$ to give the boiling point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ solution,
• The boiling point of is $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ solution $\text{100}\text{.087°C}$

To calculate the freezing point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ solution

$\text{Molal freezing-point depression constant}\text{of water is 1}\text{.86}°\text{C/molal}$

                                $\begin{array}{l}{\text{van't Hoff factor of FeCl}}_3\text{is}3.\text{4}\\ \text{ΔT=}3.\text{4×1}\text{.86 °C/molal×}\text{0}\text{.050}\text{molal}\\ \text{=}\text{0}\text{.32°C}\\ {\text{T}}_{\text{f}}\text{=}\text{0}\text{.00°C-0}\text{.32°C}\\ \text{ =-}\text{0}\text{.32°C}\end{array}$

• The given values are plugging in to equation 1 to give depression in freezing point this was subtract from freezing point of (solvent) water $\text{(100°C)}$ to give the boiling point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ solution.
• The freezing point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ solution is $-32.\text{0°C}$

Conclusion

The freezing point and the boiling point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ and $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ was calculated by using calculated elevation of boiling point and depression in freezing point and the freezing point and the boiling point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ and $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ are found to be,

• The boiling point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ is $\text{100}\text{.069°C}$
• The freezing point of $\text{0}\text{.050}\text{m}{\text{MgCl}}_{\text{2}}$ is $\text{-0}\text{.25°C}$
• The boiling point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ is $\text{100}\text{.087°C}$
• The freezing point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ is $\text{-0}\text{.32°C}$
• The freezing point of $\text{0}\text{.050}\text{m}{\text{FeCl}}_{\text{3}}$ is $\text{100}\text{.10°C}$