   Chapter 12, Problem 12.26QP General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

Solutions

Chapter
Section General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

Equal numbers of moles of two soluble, substances, substance A and substance B, are placed into separate 1.0-L samples of water. 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. 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? c Calculate the molality of the solutions of A and B. Assume that i = 1 for substance A. 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. e What concentration (molality) of substances A and B would result in both solutions having a freezing point of −0.25°C? 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.

(a)

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate 1.0 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, ΔTf = Kfm .

Where,

ΔTf = depression in freezing pointKf  = cryoscopic constantm    = molality of the solution

Explanation

While dissolving in solvent the solute particles don't dissolve in same way always...

(b)

Interpretation Introduction

Interpretation:

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

500 mL of the solution B is poured into different beaker.  The difference between the freezing point of this 500 mL solution B and 1.0 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, ΔTf = Kfm .

Where,

ΔTf = depression in freezing pointKf  = cryoscopic constantm    = molality of the solution

(c)

Interpretation Introduction

Interpretation:

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

Molality of the solutions A and B has to be calculated by assuming i = 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,

Molality = number of moles of solutemass 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, ΔTf = Kfm

(d)

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate 1.0 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,

Molality = number of moles of solutemass 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, ΔTf = Kfm

(e)

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate 1.0 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,

Molality = number of moles of solutemass 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, ΔTf = Kfm

(f)

Interpretation Introduction

Interpretation:

A and B are two substances having equal number of moles.  They are placed into separate 1.0 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,

ΔTb = iKbm .

Where,

ΔTb = elevation of boiling pointKb  = ebullioscopic constantm    = molality of the solution; i = Van't Hoff factor

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,

PA = PA° XA

Where,

PA = Partial vapor pressure of solvent in solutionPA° = Vapor pressure of pure solventXA= mole fraction of solvent in the solution

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

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

ΔP = PA° XB

Where,

XB 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,

π = MRT

Where,

π = osmotic pressureM = Molar concentrationR =  Universal gas constantT =   Temperature

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