   Chapter 11, Problem 88E

Chapter
Section
Textbook Problem

# The freezing-point depression of a 0.091-m solution of CsCl is 0.320°C. The freezing-point depression of a 0.091-m solution of CaCl2 is 0.440°C. In which solution does ion association appear to be greater? Explain.

Interpretation Introduction

Interpretation: The given two solutions does ions association appears which one is greater have to be explained.

Concept introduction:

• Elevation of boiling point:

The boiling point of the solution is increases when the solute is dissolved in the solvent are called Elevation of boiling point. It is one of the colligative Properties thus,

ΔT=iKbmsolute......(1)ΔT is boiling-point elevationKbismolal boiling-point elevation constantmis molality of the soluteiisthevan't Hoff factor

• Depression in freezing point:

The freezing point the solution is decreases when the solute is dissolved in the solvent is called depression in freezing point. it is one of the colligative Properties thus,

ΔT=iKfmsolute......(2)ΔT is boiling-point elevationKfismolal freezing-point depression constantmis molality of the soluteiisthevan't Hoff factor

Explanation

Record the given data,

The freezing-point depression of CaCl2 solution= 0.440°C

The Molality of CaCl2 solution= 0.091molal

Molal freezing-point depression constant = 1.86°C/molal

To calculate the ions association of CaCl2 solution.

Molal boiling-point elevation constant of water is 0.51 °C/molal .

ForCaCl2=i=ΔTfKfm=0.44°C1.86°C/molal×0.091molal=2.6PercentCaCl2ionized=2.6-1.03.0-1.0×100=80%20%ionassociationoccurs

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