Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Question
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Chapter 2, Problem 2.17QP

(a)

Interpretation Introduction

Interpretation: The pair in which the heavier isotope has more abundance is to be identified.

Concept introduction: The natural abundance of an isotope is calculated from the formula,

Averageatomicmass=m1×x1+m2×x2+m3×x3

To determine: If the heavier isotope has more abundance among the given pair of isotopes.

(a)

Expert Solution
Check Mark

Answer to Problem 2.17QP

Solution

The heavier isotope 11B has more abundance.

Explanation of Solution

Explanation

The natural abundance of an isotope is calculated by the formula,

Averageatomicmass=m1×x1+m2×x2+m3×x3

Where,

  • m is the mass of an isotope.
  • x is natural abundance, expressed in decimals.

The given isotopes are 10Band11B and the average atomic mass is 10.81 atomic mass unit.

The given isotopes have mass 10and11 atomic mass unit respectively.

The natural abundance of 10B is assume as x and hence, the natural abundance of 11B becomes (100x) .

Substitute the values in above formula,

Averageatomicmass=m1×x1+m2×x2+m3×x310.81=10×x100+11×100x10010.81=10x+11(100x)1001081=10x+110011x

Simplify the above equation.

x=11001081=19%

So, the natural abundance of 11B becomes,

10019=81%

Hence, the heavier isotope 11B has more abundance.

(b)

Interpretation Introduction

To determine: The heavier isotope that has more abundance for the given pairs of isotopes.

(b)

Expert Solution
Check Mark

Answer to Problem 2.17QP

Solution

The heavier isotope 7Li has more abundance.

Explanation of Solution

Explanation

The natural abundance of an isotope is calculated by the formula,

Averageatomicmass=m1×x1+m2×x2+m3×x3

Where,

  • m is the mass of an isotope.
  • x is natural abundance, expressed in decimals.

The given isotopes are 6Liand7Li and its average atomic mass is 6.941 atomic mass unit.

So, the given isotopes have mass 6and7 atomic mass unit respectively.

The natural abundance of 6Li is assume as x and hence, the natural abundance of 7Li becomes (100x) .

Substitute the values in above formula,

Averageatomicmass=m1×x1+m2×x2+m3×x36.941=6×x100+7×100x1006.941=6x+7(100x)100694.1=6x+7007x

Simplify the above equation.

x=700694.1=5.9%

So, the natural abundance of 7Li becomes,

1005.9=94.1%

Hence, the heavier isotope 7Li has more abundance.

(c)

Interpretation Introduction

To determine: The heavier isotope that has more abundance for the given pairs of isotopes.

(c)

Expert Solution
Check Mark

Answer to Problem 2.17QP

Solution

The heavier isotope 15N has not more abundance.

Explanation of Solution

Explanation

The natural abundance of an isotope is calculated by the formula,

Averageatomicmass=m1×x1+m2×x2+m3×x3

Where,

  • m is the mass of an isotope.
  • x is natural abundance, expressed in decimals.

The given isotopes are 14Nand15N and its average atomic mass is 14.01 atomic mass unit.

So, the given isotopes have mass 14and15 atomic mass unit respectively.

The natural abundance of 14N is assume as x and hence, the natural abundance of 15N becomes (100x) .

Substitute the values in above formula,

Averageatomicmass=m1×x1+m2×x2+m3×x314.01=14×x100+15×100x10014.01=14x+15(100x)1001401=14x+150015x

Simplify the above equation

x=15001401=99%

So, the natural abundance of 15N becomes,

10099=1%

Hence, the heavier isotope 15N has not more abundance.

(d)

Interpretation Introduction

To determine: The heavier isotope that has more abundance for the given pairs of isotopes.

(d)

Expert Solution
Check Mark

Answer to Problem 2.17QP

Solution

The heavier isotope 22Ne has not more abundance.

Explanation of Solution

Explanation

The natural abundance of an isotope is calculated by the formula,

Averageatomicmass=m1×x1+m2×x2+m3×x3

Where,

  • m is the mass of an isotope.
  • x is natural abundance, expressed in decimals.

The given isotopes are 20Neand22Ne and its average atomic mass is 20.18 atomic mass unit.

So, the given isotopes have mass 20and22 atomic mass unit respectively.

The natural abundance of 20Ne is assume as x and hence, the natural abundance of 22Ne becomes (100x) .

Substitute the values in above formula,

Averageatomicmass=m1×x1+m2×x2+m3×x320.18=20×x100+22×100x10020.18=20x+22(100x)1002018=20x+220022x

Simplify the above equation.

2x=220020182x=182x=1822x=91%

So, the natural abundance of 22Ne becomes,

10091=9%

Hence, the heavier isotope 22Ne has not more abundance.

Conclusion

The pair in which the heavier isotope has more abundance has been rightfully identified.

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Chapter 2 Solutions

Chemistry

Ch. 2 - Prob. 2.3VPCh. 2 - Prob. 2.4VPCh. 2 - Prob. 2.5VPCh. 2 - Prob. 2.6VPCh. 2 - Prob. 2.7VPCh. 2 - Prob. 2.8VPCh. 2 - Prob. 2.9QPCh. 2 - Prob. 2.10QPCh. 2 - Prob. 2.11QPCh. 2 - Prob. 2.12QPCh. 2 - Prob. 2.13QPCh. 2 - Prob. 2.14QPCh. 2 - Prob. 2.15QPCh. 2 - Prob. 2.16QPCh. 2 - Prob. 2.17QPCh. 2 - Prob. 2.18QPCh. 2 - Prob. 2.19QPCh. 2 - Prob. 2.20QPCh. 2 - Prob. 2.21QPCh. 2 - Prob. 2.22QPCh. 2 - Prob. 2.23QPCh. 2 - Prob. 2.24QPCh. 2 - Prob. 2.25QPCh. 2 - Prob. 2.26QPCh. 2 - Prob. 2.27QPCh. 2 - Prob. 2.28QPCh. 2 - Prob. 2.29QPCh. 2 - Prob. 2.30QPCh. 2 - Prob. 2.31QPCh. 2 - Prob. 2.32QPCh. 2 - Prob. 2.33QPCh. 2 - Prob. 2.34QPCh. 2 - Prob. 2.35QPCh. 2 - Prob. 2.36QPCh. 2 - Prob. 2.38QPCh. 2 - Prob. 2.39QPCh. 2 - Prob. 2.40QPCh. 2 - Prob. 2.41QPCh. 2 - Prob. 2.42QPCh. 2 - Prob. 2.43QPCh. 2 - Prob. 2.44QPCh. 2 - Prob. 2.45QPCh. 2 - Prob. 2.46QPCh. 2 - Prob. 2.47QPCh. 2 - Prob. 2.48QPCh. 2 - Prob. 2.49QPCh. 2 - Prob. 2.50QPCh. 2 - Prob. 2.51QPCh. 2 - Prob. 2.52QPCh. 2 - Prob. 2.53QPCh. 2 - Prob. 2.54QPCh. 2 - Prob. 2.55QPCh. 2 - Prob. 2.56QPCh. 2 - Prob. 2.57QPCh. 2 - Prob. 2.58QPCh. 2 - Prob. 2.59QPCh. 2 - Prob. 2.60QPCh. 2 - Prob. 2.61QPCh. 2 - Prob. 2.62QPCh. 2 - Prob. 2.63QPCh. 2 - Prob. 2.64QPCh. 2 - Prob. 2.65QPCh. 2 - Prob. 2.66QPCh. 2 - Prob. 2.67QPCh. 2 - Prob. 2.68QPCh. 2 - Prob. 2.69QPCh. 2 - Prob. 2.70QPCh. 2 - Prob. 2.71QPCh. 2 - Prob. 2.72QPCh. 2 - Prob. 2.73QPCh. 2 - Prob. 2.74QPCh. 2 - Prob. 2.75QPCh. 2 - Prob. 2.76QPCh. 2 - Prob. 2.77QPCh. 2 - Prob. 2.78QPCh. 2 - Prob. 2.79QPCh. 2 - Prob. 2.80QPCh. 2 - Prob. 2.81QPCh. 2 - Prob. 2.82QPCh. 2 - Prob. 2.83QPCh. 2 - Prob. 2.84QPCh. 2 - Prob. 2.85QPCh. 2 - Prob. 2.86QPCh. 2 - Prob. 2.87QPCh. 2 - Prob. 2.88QPCh. 2 - Prob. 2.89QPCh. 2 - Prob. 2.90QPCh. 2 - Prob. 2.91QPCh. 2 - Prob. 2.92QPCh. 2 - Prob. 2.93QPCh. 2 - Prob. 2.94QPCh. 2 - Prob. 2.95QPCh. 2 - Prob. 2.96QPCh. 2 - Prob. 2.97QPCh. 2 - Prob. 2.98QPCh. 2 - Prob. 2.99QPCh. 2 - Prob. 2.100QPCh. 2 - Prob. 2.101QPCh. 2 - Prob. 2.102QPCh. 2 - Prob. 2.103QPCh. 2 - Prob. 2.104APCh. 2 - Prob. 2.105APCh. 2 - Prob. 2.106APCh. 2 - Prob. 2.107APCh. 2 - Prob. 2.108APCh. 2 - Prob. 2.109APCh. 2 - Prob. 2.110APCh. 2 - Prob. 2.111APCh. 2 - Prob. 2.112APCh. 2 - Prob. 2.113APCh. 2 - Prob. 2.114APCh. 2 - Prob. 2.115APCh. 2 - Prob. 2.116APCh. 2 - Prob. 2.117APCh. 2 - Prob. 2.118APCh. 2 - Prob. 2.119APCh. 2 - Prob. 2.120APCh. 2 - Prob. 2.121APCh. 2 - Prob. 2.122APCh. 2 - Prob. 2.123APCh. 2 - Prob. 2.125APCh. 2 - Prob. 2.126APCh. 2 - Prob. 2.127AP
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