   # In the winter of 1994, record low temperatures were registered throughout the United States. For example, in Champaign, Illinois, a record low of −29°F was registered. At this temperature can salting icy roads with CaCl 2 be effective in melting the ice? a. Assume i = 3.00 for CaCl 2 . b. Assume the average value of i from Exercise 87. (The solubility of CaCl 2 in cold water is 74.5 g per 100.0 g water.) ### 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 89E
Textbook Problem
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## In the winter of 1994, record low temperatures were registered throughout the United States. For example, in Champaign, Illinois, a record low of −29°F was registered. At this temperature can salting icy roads with CaCl2 be effective in melting the ice?a. Assume i = 3.00 for CaCl2.b. Assume the average value of i from Exercise 87.(The solubility of CaCl2 in cold water is 74.5 g per 100.0 g water.)

Interpretation Introduction

Interpretation: The depression of freezing point of given solution has to be calculated.

Concept introduction:

• 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)ΔTf is freezing-point depressionKfismolal freezing-point depression constantmis molality of the soluteiisthevan't Hoff factor

### Explanation of Solution

Record the given data,

Weight of Calcium chloride = 74.5g

Weight of water = 100g

Molecular weight of Calcium chloride = 110.98g/mole

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

To find the molality of CaCl2 solution.

Tc=5(TF-32)9=55(-29-32)9=-34°C

Assuming the solubility of CaCl2 is temperature independent, the molality of a saturated CaCl2 solution is:

74.5gCaCl2100gH2O×1000gkg×1molCaCl2110.98gCaCl2=6.71molCaCl2kgH2O

Record the given data from question and step-2

Molality of CaCl2 solution= 6.71mol/kg

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

van't Hoff factor of CaCl2

i=3

To find the depression of freezing point

ΔTf=iKfm=3.00×1.86°Ckg/mol×6.71mol/kg=37

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