Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.153P

(a)

Interpretation Introduction

Interpretation:

The different molecular masses obtained for PCl3 is to be determined.

Concept introduction:

The isotopes can be defined as the atom of the same element with same number different number of neutrons.

The molar mass is defined as the mass of the all the atoms are present in the molecules in grams per mole.

(a)

Expert Solution
Check Mark

Answer to Problem 5.153P

The different molecular masses of PCl3 are 136, 138,140 and 142.

Explanation of Solution

The molecular mass of PCl3 vary due to various combinations of the two isotopes of Cl as 35Cl and 37Cl with P. The molecule PCl3 is formed by the one P atom with three Cl atoms.

The atomic mass of 35Cl isotope of chlorine is 35. If all the three chlorine present in PCl3 molecule are 35Cl then molecular mass of PCl3 is equal to the sum of atomic mass three 35Cl atoms and atomic mass of P atom. The molecular mass of PCl3 is 136.

The molecular mass of PCl3 molecule formed by the combination of a phosphorous atom, 2 35Cl atoms, and one 37Cl atom is equal to the sum of the atomic mass of 2 atoms of 35Cl isotope, one atom of 37Cl isotope and atomic mass of P atom. The molecular mass of PCl3 is 138.

The molecular mass of PCl3 molecule formed by the combination of a phosphorous atom, 1 35Cl atom, and two 37Cl atoms is equal to the sum of the atomic mass of 1 atom of 35Cl isotope, two atoms of 37Cl isotope and atomic mass of P atom. The molecular mass of PCl3 is 140.

The molecular mass of PCl3 molecule formed by a combination of the phosphorous atom and three 37Cl atoms is equal to the sum of atomic mass three atoms of 37Cl isotope and atomic mass of P atom. The molecular mass of PCl3 is 142.

The table is as follows:

Pfirst Clsecond Clthird Cltotal31353535136313735351383137373514031373737142

Conclusion

The molecular mass of PCl3 is different for a different combination of chlorine isotopes attached.

(b)

Interpretation Introduction

Interpretation:

Among the different combination of PCl3 molecules the most abundant molecule is to be determined.

Concept introduction:

The expression to calculate the fraction abundance of each isotope is as follows:

  fractional abundance of the isotope=(Percent abundance of isotope(%)100 %)

(b)

Expert Solution
Check Mark

Answer to Problem 5.153P

The most abundant PCl3 molecule that has molecular mass 136 is most abundant.

Explanation of Solution

The expression to calculate the fraction abundance of 35Cl isotope is as follows:

  The fraction abundance of 35Cl=Perecentage abundace of 35Cl100 %        (1)

Substitute 75 % for percentage abundace of 35Cl in equation (1).

  The fraction abundance of 35Cl=75 %100 %=0.75 %

The expression to calculate the fraction abundance of 37Cl is as follows:

  The fraction abundance of 37Cl=Perecentage abundance of 37Cl100 %        (2)

Substitute 25 % for the percentage of 35Cl in the equation (2).

  The fraction abundance of 37Cl=25 %100 %=0.25

The relative amount of each mass comes from the product of the relative abundance of each Cl isotope.

The expression to calculate the relative mass of each Cl isotope is as follows:

  Cl of mass 136=[(fraction abundance of  first 35Cl )(fraction abundance of  second 35Cl)(fraction abundance of third 35Cl)]        (3)

Substitute 0.75 for fraction abundance of first 35Cl, 0.75 for fraction abundance of second 35Cl and 0.75 for fraction abundance of third 35Cl in the equation (3).

  Cl of mass 136=[(0.75 )(0.75 )(0.75 )]=0.421875=0.42

The expression to calculate the relative mass of each Cl isotope is as follows:

  Cl of mass 138=[(fraction abundance of  first 37Cl )(fraction abundance of  second 35Cl)(fraction abundance of third 35Cl)]        (4)

Substitute 0.25 for fraction abundance of first 37Cl, 0.75 for fraction abundance of second 35Cl and 0.75 for fraction abundance of third 35Cl in the equation (4).

  Cl of mass 138=[(0.25 )(0.75 )(0.75 )]=0.140625

The expression to calculate the relative mass of each Cl isotope is as follows:

  Cl of mass 140=[(fraction abundance of  first 37Cl )(fraction abundance of  second 37Cl)(fraction abundance of third 35Cl)]        (5)

Substitute 0.25 for fraction abundance of first 35Cl, 0.25 for fraction abundance of second 35Cl and 0.75 for fraction abundance of third 35Cl in the equation (5).

  Cl of mass 140=[(0.25 )(0.25 )(0.75 )]=0.046875

The expression to calculate the relative mass of each Cl isotope is as follows:

  Cl of mass 142=[(fraction abundance of  first 37Cl )(fraction abundance of  second 37Cl)(fraction abundance of third 37Cl)]        (6)

Substitute 0.25 for fraction abundance of first 35Cl, 0.25 for fraction abundance of second 35Cl and 0.25 for fraction abundance of third 35Cl in the equation (6).

  Cl of mass 142=[(0.25 )(0.25 )(0.25 )]=0.015625.

Conclusion

PCl3 molecule formed by the combination of three 35Cl chlorine atoms is most abundant.

(c)

Interpretation Introduction

Interpretation:

The ratio of effusion rates of the heaviest and lightest PCl3 molecule is to be calculated.

Concept introduction:

Effusion is explained as the movement of the gas molecule through a pinhole.

Diffusion can be explained as the mixing of one gas molecule with another gas molecule by random motion.

According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

  Rate of ARate of B=urms of Aurms of B=MBMA=time Btime A

Here,

MB is the molar mass of gas B

MA is the molar mass of gas A.

(c)

Expert Solution
Check Mark

Answer to Problem 5.153P

The ratio of effusion rates of the heaviest and lightest PCl3 molecule is 0.979.

Explanation of Solution

The heaviest PCl3 molecule is P37Cl3 and the lightest PCl3 molecule is P35Cl3.

The expression to calculate the ratio of the rate of the heaviest and lightest PCl3 molecule is as follows:

  Rate of P37Cl3Rate of P35Cl3=Molar mass of P35Cl3Molar mass of P37Cl3        (7)

Substitute 136g/mol for P35Cl3 and 142g/mol for P37Cl3 in the equation (7).

  Rate of 37ClRate of 35Cl=136g/mol142g/mol=0.9786450.979.

Conclusion

The ratio of effusion rates of the heaviest and lightest PCl5 molecule is less than 1 this suggests that the effusion rate of P35Cl3 is greater than P37Cl3.

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Chapter 5 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

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