   # Calculate the freezing point and the boiling point of each of the following aqueous solutions. (Assume complete dissociation.) a. 0.050 m MgCl 2 b. 0.050 m FeCl 3 ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 10, Problem 85E
Textbook Problem
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## Calculate the freezing point and the boiling point of each of the following aqueous solutions. (Assume complete dissociation.)a. 0.050 m MgCl2b. 0.050 m FeCl3

Interpretation Introduction

Interpretation: The freezing point and the boiling point of the given solutions are should be calculated.

Concept introduction:

• Elevation of boiling point:

The boiling point of the solution is increases when the solute is dissolved in the solvent is 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 of Solution

Record the given data,

Molality of MgCl2= 0.050m

Molality of FeCl3=0.050m

To calculate the boiling point of 0.050mMgCl2 solution.

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

i=moleofionsmoleofsolutei=3.0/1ΔT=3.0×0.51 °C/molal×0.050molal=0.077°C=100°C+0.077°CTb=100.077°C

• The given values are plugging in to equation 1 to give elevation of boiling point this was added to boiling point of (solvent) water (100°C) to give the boiling point of 0.050mMgCl2 solution.
• The boiling point of 0.050mMgCl2 solution is 100.077°C

To calculate the freezing point of 0.050mMgCl2 solution

Molal freezing-point depression constantof water is 1.86°C/molal

i=moleofionsmoleofsolutei=3.0/1ΔT=3.0×1.86 °C/molal×0.050molal=0.28°CTf=0.00°C-0.28°C=-28.0°C

• The given values are plugging in to equation 1 to give depression in freezing point this was subtract from freezing point of (solvent) water (100°C) to give the boiling point of 0.050mMgCl2 solution.
• The freezing point of 0

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