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.)

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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 = iK f m solute ...... ( 2 ) ΔT f is freezing-point depression K f is molal freezing-point depression constant m is molality of the solute i is the van't Hoff factor Explanation Record the given data, Weight of Calcium chloride = 74 .5 g Weight of water = 100 g Molecular weight of Calcium chloride = 110 .98 g/mole Molal freezing-point depression constant = 1 .86°C/molal To find the molality of CaCl 2 solution. Tc= 5(T F -32) 9 =5 5(-29-32) 9 =-34°C Assuming the solubility of CaCl 2 is temperature independent, the molality of a saturated CaCl 2 solution is: 74 .5 g CaCl 2 100 g H 2 O × 1000 g kg × 1 mol CaCl 2 110 .98 g CaCl 2 = 6 .71 mol CaCl 2 kg H 2 O Record the given data from question and step-2 Molality of CaCl 2 solution= 6 .71 mol /kg Molal freezing-point depression constant = 1 .86°C/molal van't Hoff factor of CaCl 2 i=3 To find the depression of freezing point ΔT f =iK f m=3 .00×1 .86°C kg/mol×6 .71 mol/kg=37

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Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach

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Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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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 depression Kf is molal freezing-point depression constant m is molality of the solute i is the van't Hoff factor

Explanation of Solution

Record the given data,

Weight of Calcium chloride = 74.5 g

Weight of water = 100 g

Molecular weight of Calcium chloride = 110.98 g/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:

74.5 g CaCl2100 g H2O×1000 gkg×1 mol CaCl2110.98 g CaCl2=6.71 mol CaCl2kg H2O

Record the given data from question and step-2

Molality of CaCl2 solution= 6.71 mol /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°C kg/mol×6.71 mol/kg=37

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