EBK PHYSICAL CHEMISTRY
EBK PHYSICAL CHEMISTRY
2nd Edition
ISBN: 8220100477560
Author: Ball
Publisher: Cengage Learning US
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Chapter 7, Problem 7.44E
Interpretation Introduction

Interpretation:

The approximate molarity of the given solution and the value of Henry’s law constant for the given gas are to be calculated.

Concept introduction:

According to Henry’s law, the temperature and solubility of dissolved gas are inversely related. When temperature increases, kinetic energy increases which causes the molecules to break the intermolecular bonds and escape from the solution. Hence solubility decreases.

Expert Solution & Answer
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Answer to Problem 7.44E

The approximate molarity is 2.32×103 M. The Henry’s law constant is 2.43×109 Pa.

Explanation of Solution

The mole fraction of CCl2F2 was found to be 4.17×105 at normal pressure.

The mass of CCl2F2 is calculated by the formula,

Mass of CCl2F2=Numberofmoles×Molar mass of CCl2F2

Molar mass of CCl2F2 is 120.91 g/mol.

The number of moles of CCl2F2 is 4.17×105.

The number of moles of H2O is 1(4.17×105)=0.999

Substitute the values of mass and molar mass in the above formula.

Mass of CCl2F2=Numberofmoles×Molar mass of CCl2F2=4.17×10-5 mol×120.91 g/mol=5.04×10-3 g

The mass of CCl2F2 is calculated by the formula,

Mass of CCl2F2=Numberofmoles×Molar mass of CCl2F2

Molar mass of CCl2F2 is 120.91 g/mol.

The number of moles of CCl2F2 is 4.17×105.

Substitute the values of mass and molar mass in the above formula.

Mass of CCl2F2=Numberofmoles×Molar mass of CCl2F2=4.17×10-5 mol×120.91 g/mol=5.04×10-3 g

The mass of H2O is calculated by the formula,

Mass of H2O=Numberofmoles×Molar mass of H2O

Molar mass of H2O is 18 g/mol.

Substitute the values of mass and molar mass of H2O in the above formula.

Mass of H2O=Numberofmoles×Molar mass of H2O=0.999 mol×18 g/mol=17.982 g

The mass of the solution will be,

Mass of solution=Mass of CCl2F2+Mass of H2O=5.04×10-3 g+17.982 g=17.987 g

The volume of solution is calculated by the formula,

Volume=MassofsolutionDensity of water

The density of water is 1.00 g/cm3.

Substitute the value of mass and density in the above formula.

Volume=17.987g1 g/cm3=17.987 cm3

Convert 17.987 cm3 to L.

17.987 cm3=17.981000L=0.0179L

Therefore, the volume of solution is 0.0179L.

The molarity of the aqueous solution is calculated by the formula,

Molarity=Moles of SoluteLiters of solution

Substitute the values of number of moles of CCl2F2 and volume in the above equation.

Molarity=Moles of SoluteLiters of solution=4.17×105 mol0.0179 L=2.32×103 mol/L=2.32×103 M

The expression for Henry’s Law is shown below.

pi=Ki×xi

Where,

pi is the pressure.

Ki is Henry constant.

xi is mole fraction.

The standard pressure in Pascal’s is 101325 Pa.

The value of mole fraction is 4.17×105.

Substitute the value of pressure and value of mole fraction in the above expression.

KCCl2F2=pxCCl2F2=101325 Pa4.17×105=2.43×109 Pa

Therefore, the Henry’s law constant is 2.43×109 Pa.

Conclusion

The Henry’s law constant is 2.43×109 Pa.

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Chapter 7 Solutions

EBK PHYSICAL CHEMISTRY

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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY