Chapter 11, Problem 115AE

Interpretation Introduction

Interpretation: The boiling point and vapor pressure of the given aqueous 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 depression in freezing point. it is one of the colligative Properties thus,

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

• Raoult's law:

  The mole fraction of a solute is related to the vapor pressure of the solution thus,

                                 $\begin{array}{l}{\text{P}}_{\text{solution}}{\text{=P°}}_{\text{solvent}}{\text{X}}_{\text{solvent}}\mathrm{......}\text{(3)}\\ \text{P}\text{is}\text{vapor pressure}\text{of the solution}\\ {\text{P°}}_{\text{solvent}}\text{pressure}\text{of the solvent}\\ {\text{X}}_{\text{solvent}}\text{ mole fraction of}\text{solvent}\end{array}$

Answer to Problem 115AE

The boiling point solution is $\text{100}\text{.77°C}$.

The vapor pressure of the given aqueous solution is $\text{23}\text{.1}\text{mm}\text{Hg}$.

Explanation of Solution

Record the given data,

           Freezing point of an aqueous solution $\text{=- 2}\text{.79°C}$

To calculate molality of the aqueous solution:

           $\begin{array}{l}\text{Molal boiling-point elevation constant}\text{of water =}\text{0}\text{.51}°\text{C/mol}\\ \text{Molal freezing-point elevation constant}\text{of water}\text{=}\text{1}\text{.86}°\text{C/mol}\end{array}$

                                       $\begin{array}{l}\text{m}\text{=}\frac{{\text{ΔT}}_{\text{f}}}{{\text{K}}_{\text{f}}}\\ \text{=}\frac{\text{2}\text{.79°C}}{\text{1}\text{.86°C/molal}}\\ \text{=}\text{1}\text{.50}\text{molal}\end{array}$

• The given values are plugging in to equation 1 and do some rearrangement to give molality of the aqueous solution.
• The molality of the aqueous solution is $\text{1}\text{.50}\text{molal}$.

To calculate the boiling point of aqueous solution.

                                     $\begin{array}{l}\text{ΔT}\text{=}{\text{K}}_{\text{b}}\text{m}\\ \text{=}\text{0}\text{.51°C/molal×1}\text{.50}\text{molal}\\ \text{=}\text{0}\text{.77°C}\\ \text{=100+}\text{0}\text{.77°C}\\ \text{=}\text{100}\text{.077°C}\\ \end{array}$

• The calculated molality of the aqueous solution and molal boiling point elevation constant are plugging in to equation 1and do some rearrangement to give boiling point of the aqueous solution.
• The boiling point of aqueous solution is $\text{100}\text{.077°C}$.

Record the given data,

Vapor pressure of pure water at $\text{25°C}\text{=23}\text{.76 mm Hg}$

To calculate the vapor pressure of the given aqueous solution.

                                 $\begin{array}{l}{\text{χ}}_{\text{water}}\text{=}\frac{\text{mol}\text{water}}{\text{mol}\text{of}\text{solution}}\\ \text{Assuming}\text{1}\text{.50}\text{mol}\text{of}\text{solute}\text{dissolve}\text{in}\text{1}\text{kg}\text{of}\text{water}\\ \text{mol}\text{water}\text{=}\text{1}{\text{.00×10}}^{\text{3}}\text{g}\text{×}\frac{\text{1}\text{mol}\text{water}}{\text{18}\text{.02}\text{g}\text{water}}\text{=}\text{55}\text{.5}\text{mol}\\ {\text{χ}}_{\text{water}}\text{=}\frac{\text{55}\text{.5}\text{mol}}{\text{1}\text{.5+}\text{55}\text{.5}}\\ \text{=}\text{0}\text{.974}\\ {\text{P}}_{\text{sol}}\text{=}\text{0}\text{.974×23}\text{.76mm}\text{Hg}\\ \text{=}\text{23}\text{.1}\text{mm}\text{Hg}\\ \end{array}$

• The mole fraction of water is calculated by the assumed volume and molecular weight are plugging in to above equation to give mole fraction of water.
• The calculated mole fraction and vapor pressure of water are plugging equation 3 to give vapor pressure of the given aqueous solution.
• The vapor pressure of the given aqueous solution is $\text{23}\text{.1}\text{mm}\text{Hg}$.

To explain: the necessary assumption of solving above problems.

• The assumption are solution is in ideal, solute was nonvolatile so the van't Hoff factor values are assumed to be 1 in to solving in above problems a and b.
• The solute is not ionic and it is not formed any ions in solution.

Conclusion

• The boiling point of given aqueous solution was calculated by using calculated molality and molal boiling point elevation constant and the boiling point of aqueous solution was found to be $\text{100}\text{.077°C}$.
• The vapor pressure of the given aqueous solution was calculated by using mole fraction and vapor pressure of water and the vapor pressure of the given aqueous solution was found to be $\text{23}\text{.1}\text{mm}\text{Hg}$.