Chapter 12, Problem 12.26QP

Equal numbers of moles of two soluble, substances, substance A and substance B, are placed into separate 1.0-L samples of water.

1. a The water samples are cooled. Sample A freezes at −0.50°C, and Sample B freezes at −l.00°C. Explain how the solutions can have different freezing points.
2. b You pour 500 mL of the solution containing substance B into a different beaker. How would the freezing point of this 500-mL portion of solution B compare to the freezing point of the 1.0-L sample of solution A?
3. c Calculate the molality of the solutions of A and B. Assume that i = 1 for substance A.
4. d If you were to add an additional 1.0 kg of water to solution B, what would be the new freezing point of the solution? Try to write an answer to this question without using a mathematical formula.
5. e What concentration (molality) of substances A and B would result in both solutions having a freezing point of −0.25°C?
6. f Compare the boiling points, vapor pressure, and osmotic pressure of the original solutions of A and B. Don’t perform the calculations; just state which is the greater in each ease.

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate $1.0\text{ L}$ samples of water.  The difference in freezing point of the substances A and B has to be explained.

Concept Introduction:

Freezing point of the substance is temperature at which liquid substance remains in equilibrium with solid substance.

Depression in freezing point of a substance can be determined by the formula, $\Delta T_f{\text{ = K}}_f\text{m}$.

Where,

$\begin{array}{l}\Delta T_f\text{ = depression in freezing point}\\ {\text{K}}_f\text{ = cryoscopic constant}\\ \text{m = molality of the solution}\end{array}$

Explanation of Solution

While dissolving in solvent the solute particles don't dissolve in same way always.  This is why, even though the two sample solutions A and B are made of same type of solute and have equal number of moles has different freezing point.

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate $1.0\text{ L}$ samples of water.

$500\text{ mL}$ of the solution B is poured into different beaker.  The difference between the freezing point of this $500\text{ mL}$ solution B and $1.0\text{ L}$ sample of solution A has to be explained.

Concept Introduction:

Freezing point of the substance is temperature at which liquid substance remains in equilibrium with solid substance.

Depression in freezing point of a substance can be determined by the formula, $\Delta T_f{\text{ = K}}_f\text{m}$.

Where,

$\begin{array}{l}\Delta T_f\text{ = depression in freezing point}\\ {\text{K}}_f\text{ = cryoscopic constant}\\ \text{m = molality of the solution}\end{array}$

Explanation of Solution

Freezing point of a solution is a colligative property which depends only on the concentration of the solute.  Thus dividing the volume of solution of same concentration doesn't result in variation of freezing point.  So there will be no change in difference between the boiling point of $500\text{ mL}$ of solution B and $1.0\text{ L}$ of solution A.

Conclusion

Two different solutions made of same solute with same number of moles can have different freezing points.  Freezing point of a solution doesn’t vary with volume and it varies only with concentration of the solute in the solution.

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate $1.0\text{ L}$ samples of water.

Molality of the solutions A and B has to be calculated by assuming $i\text{ = 1}$ for substance A.

Concept Introduction:

Molality is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

$\text{Molality = }\frac{number\text{ of }molesof\text{ solute}}{mass\text{ of solvent in kg}}$

Depression in freezing point of a substance can also be used to determine molality of the solution by rewriting the formula, $\Delta T_f{\text{ = K}}_f\text{m}$

Answer to Problem 12.26QP

Molality of the solutions A and B are calculated to be $0.27m.$

Explanation of Solution

Both the solutions A and B have same molality as both of them have same number of moles of solute.  As given $i=1$, and freezing point of the solution, molality of both the solutions A and B are calculated to be,

$\begin{array}{l}{c}_m\text{= }\frac{\Delta T_f}{K_f}\\ =\frac{0.50°C}{1.858°C/m}\\ =\text{ 0}\text{.27 m}\end{array}$

Conclusion

Since depression in freezing point depends upon the concentration of the solute, this parameter can be used to determine the molal concentration of the solution.

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate $1.0\text{ L}$ samples of water.  The molality at which both the solutions A and B would have same freezing point of $-0.25°C$ has to be determined.

Concept Introduction:

Molality is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

$\text{Molality = }\frac{number\text{ of }molesof\text{ solute}}{mass\text{ of solvent in kg}}$

Depression in freezing point of a substance can also be used to determine molality of the solution by rewriting the formula, $\Delta T_f{\text{ = K}}_f\text{m}$

Answer to Problem 12.26QP

The new freezing point will be half of the freezing point of solution B that is $-0.50°C.$

Explanation of Solution

Adding $1.0\text{ kg}$ of water to the solution B reduces the molality to one half of the actual molality.  Consequently $\Delta T_f$ will be reduced by one-half and thus the new freezing point of the solution would be $-0.50°C.$

Conclusion

Addition of excess solvent varies the solute concentration. Varying the solute concentration varies the freezing point of the solution.

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate $1.0\text{ L}$ samples of water.  Osmotic pressure, vapor pressure, boiling points of both the solutions A and B has to be compared by theoretical explanation.

Concept Introduction:

Molality is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

$\text{Molality = }\frac{number\text{ of }molesof\text{ solute}}{mass\text{ of solvent in kg}}$

Depression in freezing point of a substance can also be used to determine molality of the solution by rewriting the formula, $\Delta T_f{\text{ = K}}_f\text{m}$

Answer to Problem 12.26QP

The two solutions A and B have same freezing point at molality of $0.13m.$

Explanation of Solution

As given $i=1$, and freezing point of the solution, molality of both the solutions A and B in which they both have same freezing point, is calculated to be,

$\begin{array}{l}{c}_m\text{= }\frac{\Delta T_f}{K_f}\\ =\frac{0.25°C}{1.858°C/m}\\ =\text{ 0}\text{.13 m}\end{array}$

Conclusion

Since depression in freezing point depends upon the concentration of the solute, this parameter can be used to determine the molal concentration of the solution.

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate $1.0\text{ L}$ samples of water.

Osmotic pressure, vapor pressure, boiling points of both the solutions A and B has to be compared by theoretical explanation.

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.

Elevation in boiling point of a substance can be determined by the formula,

$\Delta T_b{\text{ = iK}}_b\text{m}$.

Where,

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

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.

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.

In the process of osmosis, the solvent molecules pass through a semi - permeable membrane from less concentrated solution to more concentrated solution.  The pressure that has to be applied to prevent the flow of solvent molecules is called osmotic pressure.  It is expressed as,

$\pi =MRT$

Where,

$\begin{array}{l}\text{π}\text{= osmotic pressure}\\ \text{M}\text{= Molar concentration}\\ \text{R}\text{= Universal gas constant}\\ \text{T}\text{= Temperature}\end{array}$

Answer to Problem 12.26QP

The solution A will have lower boiling point, lower osmotic pressure and higher vapor pressure than solution B.

Explanation of Solution

Vapor pressure and boiling point share inverse relationship. But decrease in vapor pressure leads to decrease in freezing point.  Higher the vapor pressure of the solution lower will be the boiling point of the solution.  Since solution A has lower freezing point than solution B, it indicates lower vapor pressure and higher boiling point than solution B.  Solution A has lower freezing point that it has lower osmotic pressure than solution B.

Conclusion

Vapor pressure and boiling point of the solution are dependent on each other.  Osmotic pressure of the solution can be determined on the basis of freezing point or depression in freezing point.