   Chapter 11, Problem 79E

Chapter
Section
Textbook Problem

# Consider the following solutions:0.010 m Na3PO4 in water0.020 m CaBr2 in water0.020 m KCl in water0.020 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?

a)

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,

Psolutionsolventsolvent

Where, Psolution   = observed vapor pressure of the solution

χsolvent    = mole fraction of solvent

solvent = vapor pressure of pure solvent

The freezing point depression can be calculated from the equation,

ΔT=Kmsolute

Where, ΔT =change in freezing point depression

K = molal freezing point depression/boiling point constant

msolute = molality of solute.

Explanation

To identify the solution that has the same boiling point as 0.040mC6H2O6

0.010mNa3PO4and0.020mKCl has the same boiling point as 0.040mC6H2O6 .

Na3PO4(s)3Na+(aq)+PO43-(aq)i=4.0CaBr2(s)Ca2+(aq)+2Br-(aq)i=3

b)

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,

Psolutionsolventsolvent

Where, Psolution   = observed vapor pressure of the solution

χsolvent    = mole fraction of solvent

solvent = vapor pressure of pure solvent

The freezing point depression can be calculated from the equation,

ΔT=Kmsolute

Where, ΔT =change in freezing point depression

K = molal freezing point depression/boiling point constant

msolute = molality of solute.

c)

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,

Psolutionsolventsolvent

Where, Psolution   = observed vapor pressure of the solution

χsolvent    = mole fraction of solvent

solvent = vapor pressure of pure solvent

The freezing point depression can be calculated from the equation,

ΔT=Kmsolute

Where, ΔT =change in freezing point depression

K = molal freezing point depression/boiling point constant

msolute = molality of solute.

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