Chapter 12, Problem 12.25QP

Consider two hypothetical pure substances, AB(s) and XY(s). When equal molar amounts of these substances are placed in separate 500-mL samples of water, they undergo the following reactions:

$\begin{array}{c}\text{AB}(s)\to {\text{A}}^{+}(aq)+{\text{B}}^{-}(aq)\\ \text{XY}(s)\to \text{XY}(aq)\end{array}$

1. a Which solution would you expect to have the lower boiling point? Why?
2. b Would you expect the vapor pressures of the two solutions to be equal? If not, which one would you expect to have the higher vapor pressure?
3. c Describe a procedure that would make the two solutions have the same boiling point.
4. d If you took 250 mL of the AB(aq) solution prepared above, would it have the same boiling point as the original solution? Be sure to explain your answer.
5. e The container of XY(aq) is left out on the bench top for several days, which allows some of the water to evaporate from the solution. How would the melting point of this solution compare to the melting point of the original solution?

Interpretation Introduction

Interpretation:

Two pure hypothetical substances have to be considered – XY(s) and AB(s).  The two substances are taken in equal number of moles and made into solution in separate $500mL$ beakers.  The following reactions occur –

$\begin{array}{l}AB(s)\underset{}{\to }{\text{ A}}^{+}{\text{(aq) + B}}^{-}(aq)\\ XY(s)\stackrel{}{\to }\text{ XY(aq)}\end{array}$

The solution having higher boiling point has to be identified and the same has to be explained.

Concept Introduction:

Boiling point of a liquid substance is defined as the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure.

Boiling point of a substance can be determined by the formula, $\Delta T_b{\text{ = iK}}_b\text{m}$.

Where,

$\begin{array}{l}\Delta T_b\text{ = elevation of boiling point}\\ {\text{K}}_b\text{ = ebullioscopic constant}\\ \text{m = molality of the solution; i = Van't Hoff factor}\end{array}$

Answer to Problem 12.25QP

Aqueous solution of AB will have higher boiling than that of XY.

Explanation of Solution

Boiling point of a substance can be determined by the formula $\Delta T_b{\text{ = iK}}_b\text{m}$. Greater the value of $\Delta T_b$, greater will be the boiling point of the substance.  According to the formula, $\Delta T_b$ is directly proportional to van’t Hoff factor ‘i’.  Here ‘i’ refers to the number of moles of ions furnished by the substance when it is dissolved in water.  Thus greater the value of ‘i’ for a substance, greater is its boiling point.  The two solutions have same volume but the substance undergo different reaction in solution.

Substance AB furnishes two ions in solution and so it has ‘i’ value 2.  XY dissolves as molecular compound and don’t form any ions and has ‘i’ value 1.  Hence solution of AB will have higher boiling point than that of XY.

Conclusion

Substance that produces more number of ions per unit formula tends to have higher boiling point.

Interpretation Introduction

Interpretation:

Two pure hypothetical substances have to be considered – XY(s) and AB(s).  The two substances are taken in equal number of moles and made into solution in separate $500mL$ beakers.  The following reactions occur –

$\begin{array}{l}AB(s)\underset{}{\to }{\text{ A}}^{+}{\text{(aq) + B}}^{-}(aq)\\ XY(s)\stackrel{}{\to }\text{ XY(aq)}\end{array}$

Vapor pressure of the two solutions has to be compared.

Concept Introduction:

Vapor pressure of a substance is known as the pressure exerted by molecules on the vapor phase when they are in equilibrium with their actual phase which can be liquid or solid.

A substance is said to be volatile if it vaporizes readily at room temperature itself.  Such substances have high vapor pressure as most of its molecules tend to exist in vapor phase.  A substance is said to be non-volatile if it doesn’t vaporize spontaneously and remains stable.

Vapor pressure of a volatile solvent can be lowered by addition of a non-volatile solute. Raoult’s law deals with the vapor pressure of pure solvents and solution which states –

Partial pressure of solvent is equivalent to the product of vapor pressure of the solvent in its pure state and mole fraction of solvent in the solution.  It is expressed as,

$P_A=P_A^{°}{X}_A$

Where,

$\begin{array}{l}{P}_A\text{ = Partial vapor pressure of solvent in solution}\\ {\text{P}}_A^{°}\text{ = Vapor pressure of pure solvent}\\ {\text{X}}_A\text{= mole fraction of solvent in the solution}\end{array}$

When the solute is non-volatile, the vapor pressure of the whole solution is equal to $P_A.$

The lowering of vapor pressure of the solvent due to the addition of non-volatile solute is expressed as,

$\Delta P=P_A^{°}{X}_B$

Where,

  $X_B$ is the mole fraction of the solute.

If the solute is an Ionic compound, the equation becomes,

$\Delta P=i\times P_A^{°}{X}_B$

Where,

$i=van\text{'}tHofffactor$ refers to number of ions per unit formula produced by ionic solute.

Answer to Problem 12.25QP

Aqueous solution of XY will have higher vapor pressure than that of AB.

Explanation of Solution

According to Raoult’s law, lowering of vapor pressure is directly proportional to ‘i’ value of the substance.  Substance AB furnishes two ions in solution and so it has ‘i’ value 2.  Hence the vapor pressure is lowered to more extent in this case.  XY dissolves as molecular compound and don’t form any ions and has ‘i’ value 1.  Hence solution of XY will have higher vapor pressure than that of AB.

Conclusion

Substance that produces more number of ions per unit formula tends to have lower vapor pressure.

Interpretation Introduction

Interpretation:

Two pure hypothetical substances have to be considered – XY(s) and AB(s).  The two substances are taken in equal number of moles and made into solution in separate $500mL$ beakers.  The following reactions occur –

$\begin{array}{l}AB(s)\underset{}{\to }{\text{ A}}^{+}{\text{(aq) + B}}^{-}(aq)\\ XY(s)\stackrel{}{\to }\text{ XY(aq)}\end{array}$

A procedure to make the two solutions to have same boiling point has to be described.

Explanation of Solution

As the compound AB has ‘i’ factor 2, altering the ‘i’ factor of XY from 1 to 2, results in two solutions to have same boiling point.  Doubling the concentration of XY makes XY to have ‘i’ factor 2 that eventually both the solutions will have same boiling point.

Conclusion

The ‘i’ factor of solute in the solution governs the boiling point and many other colligative properties of the substance.

Interpretation Introduction

Interpretation:

Two pure hypothetical substances have to be considered – XY(s) and AB(s).  The two substances are taken in equal number of moles and made into solution in separate $500mL$ beakers.  The following reactions occur –

$\begin{array}{l}AB(s)\underset{}{\to }{\text{ A}}^{+}{\text{(aq) + B}}^{-}(aq)\\ XY(s)\stackrel{}{\to }\text{ XY(aq)}\end{array}$

The boiling point of original solution of AB(s) and its $250mL$ aqueous solution has to be compared.

Explanation of Solution

Boiling point of any substance is independent of volume or quantity.  Hence the boiling point of aqueous solution of AB and $250mL$ aqueous solution of AB don’t vary and they are same.

Conclusion

Boiling point of a substance doesn’t vary with volume and it varies only with concentration of the solute in the solution.

Interpretation Introduction

Interpretation:

Two pure hypothetical substances have to be considered – XY(s) and AB(s).  The two substances are taken in equal number of moles and made into solution in separate $500mL$ beakers.  The following reactions occur –

$\begin{array}{l}AB(s)\underset{}{\to }{\text{ A}}^{+}{\text{(aq) + B}}^{-}(aq)\\ XY(s)\stackrel{}{\to }\text{ XY(aq)}\end{array}$

Melting point of the original solution of XY and that of the solution in which some quantity of solvent has been evaporated has to be compared.

Concept Introduction:

Melting point of a substance is the temperature at which a solid substance remains in equilibrium with liquid substance whereas freezing point of the substance is temperature at which liquid substance remains in equilibrium with solid substance.

Answer to Problem 12.25QP

Melting point (Freezing point) of original solution will be higher than that of the solution in which some amount of the solvent has been evaporated.

Explanation of Solution

Freezing point of the solution depends upon the concentration of solute.  Greater the concentration of the solute, lower will be the freezing point of the solution.  When solvent evaporates the concentration of solute XY becomes higher that it will freeze at lower temperature than that of the pure solvent and that of the original solution XY (aq).

Conclusion

Higher the solute concentration, lower will be the freezing point of solution.