Chapter 10, Problem 79E

Consider the following solutions:

0.010 m Na₃PO₄ in water

0.020 m CaBr₂ in water

0.020 m KCl in water

0.020 m HF in water (HF is a weak acid.)

a. Assuming complete dissociation of the soluble salts, which solution(s) would have the same boiling point as 0.040 m C₆H₁₂O₆ in water? C₆H₁₂O₆ is a nonelectrolyte.

b. Which solution would have the highest vapor pressure at 28°C?

c. Which solution would have the largest freezing-point depression?

Interpretation Introduction

Interpretation: Among the given set of solutions, the one with the same boiling point of Glucose, highest vapor pressure and largest freezing point depression has to be determined.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The vapor pressure of solution can be calculated from Raoult’s law,

${\text{P}}_{\text{solution}}{\text{=χ}}_{\text{solvent}}{\text{P°}}_{\text{solvent}}$

Where, ${\text{P}}_{\text{solution}}$  = observed vapor pressure of the solution

${\text{χ}}_{\text{solvent}}$   = mole fraction of solvent

            ${\text{P°}}_{\text{solvent}}$ = vapor pressure of pure solvent

The freezing point depression can be calculated from the equation,

${\text{ΔT=Km}}_{\text{solute}}$

Where, $\text{ΔT}$=change in freezing point depression

$\text{K}$= molal freezing point depression/boiling point constant

${\text{m}}_{\text{solute}}$ = molality of solute.

Answer to Problem 79E

Answer

$\text{0}\text{.010}\text{m}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\text{and}\text{0}\text{.020}\text{m}\text{KCl}$ has the same boiling point as $\text{0}\text{.040}\text{m}{\text{C}}_{\text{6}}{\text{H}}_{\text{2}}{\text{O}}_{\text{6}}$.

Explanation of Solution

To identify the solution that has the same boiling point as $\text{0}\text{.040}\text{m}{\text{C}}_{\text{6}}{\text{H}}_{\text{2}}{\text{O}}_{\text{6}}$

$\text{0}\text{.010}\text{m}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\text{and}\text{0}\text{.020}\text{m}\text{KCl}$ has the same boiling point as $\text{0}\text{.040}\text{m}{\text{C}}_{\text{6}}{\text{H}}_{\text{2}}{\text{O}}_{\text{6}}$.

$\begin{array}{l}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\left(\text{s}\right)}\to {\text{3Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+PO}}_{\text{4}}{{}^{\text{3-}}}_{\left(\text{aq}\right)}\text{i=4}\text{.0}\\ \\ {\text{CaBr}}_{\text{2(s)}}\to {\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\text{i=3}\text{.0}\\ \\ {\text{KCl}}_{\left(\text{s}\right)}\to {\text{K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\text{i=2}\text{.0}\\ \end{array}$

Assuming complete dissociation, the effective particle concentrations of solution are,

$\begin{array}{l}\text{4}\text{.0(0}\text{.010}\text{molal)=0}\text{.040}\text{molal}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\\ \\ \text{3}\text{.0(0}\text{.020}\text{molal)=0}\text{.060}\text{molal}{\text{CaBr}}_{\text{2}}\\ \\ \text{2}\text{.0(0}\text{.020}\text{molal)=0}\text{.040}\text{molal}\text{KCl}\end{array}$

Slightly greater $\text{0}\text{.020}\text{molal}\text{HF}$ solution because $\text{HF}$ is weak acid and it partially dissociates in Water.

Therefore, $\text{0}\text{.010}\text{m}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\text{and}\text{0}\text{.020}\text{m}\text{KCl}$ has the same boiling point as $\text{0}\text{.040}\text{m}{\text{C}}_{\text{6}}{\text{H}}_{\text{2}}{\text{O}}_{\text{6}}$.

Interpretation Introduction

Interpretation: Among the given set of solutions, the one with the same boiling point of Glucose, highest vapor pressure and largest freezing point depression has to be determined.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The vapor pressure of solution can be calculated from Raoult’s law,

${\text{P}}_{\text{solution}}{\text{=χ}}_{\text{solvent}}{\text{P°}}_{\text{solvent}}$

Where, ${\text{P}}_{\text{solution}}$  = observed vapor pressure of the solution

${\text{χ}}_{\text{solvent}}$   = mole fraction of solvent

            ${\text{P°}}_{\text{solvent}}$ = vapor pressure of pure solvent

The freezing point depression can be calculated from the equation,

${\text{ΔT=Km}}_{\text{solute}}$

Where, $\text{ΔT}$=change in freezing point depression

$\text{K}$= molal freezing point depression/boiling point constant

${\text{m}}_{\text{solute}}$ = molality of solute.

Answer to Problem 79E

Answer

$\text{0}\text{.020}\text{m}\text{HF}$ has the highest vapor pressure among the given solutions.

Explanation of Solution

To identify which solution has highest vapor pressure at $\text{28°C}$

$\text{0}\text{.020}\text{m}\text{HF}$ has the highest vapor pressure among the given solutions.

As the concentration of solute decreases, the vapor pressure of the solvent gets increases because the mole fraction gets increased.

Hence, $\text{0}\text{.020}\text{molal}\text{HF}$ has the highest vapor pressure since the effective particle concentration of $\text{0}\text{.020}\text{molal}\text{HF}$ is very small.

Interpretation Introduction

Interpretation: Among the given set of solutions, the one with the same boiling point of Glucose, highest vapor pressure and largest freezing point depression has to be determined.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The vapor pressure of solution can be calculated from Raoult’s law,

${\text{P}}_{\text{solution}}{\text{=χ}}_{\text{solvent}}{\text{P°}}_{\text{solvent}}$

Where, ${\text{P}}_{\text{solution}}$  = observed vapor pressure of the solution

${\text{χ}}_{\text{solvent}}$   = mole fraction of solvent

            ${\text{P°}}_{\text{solvent}}$ = vapor pressure of pure solvent

The freezing point depression can be calculated from the equation,

${\text{ΔT=Km}}_{\text{solute}}$

Where, $\text{ΔT}$=change in freezing point depression

                                        $\text{K}$= molal freezing point depression/boiling point constant

${\text{m}}_{\text{solute}}$ = molality of solute.

Answer to Problem 79E

Answer

$\text{0}\text{.020}\text{m}{\text{CaBr}}_{\text{2}}$ has the largest vapor pressure among the given solutions.

Explanation of Solution

To identify which solution has largest freezing point depression.

$\text{0}\text{.020}\text{m}{\text{CaBr}}_{\text{2}}$ has the largest vapor pressure among the given solutions.

The largest freezing point depression is seen in $\text{0}\text{.020}\text{m}{\text{CaBr}}_{\text{2}}$ because of its large effective particle concentration.