Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 10, Problem 10.59QP

(a)

Interpretation Introduction

Interpretation: The solubility of O2 in the blood of a climber on Mt. Everest and a scuba diver at 100 feet is to be calculated.

Concept introduction: The Henry’s law states that “the amount of gas dissolved in given liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.”

To determine: The solubility of O2 in the blood of a climber on Mt. Everest.

(a)

Expert Solution
Check Mark

Answer to Problem 10.59QP

Solution

The solubility of O2 in the blood of a climber on Mt. Everest is 0.0027mol/L_ .

Explanation of Solution

Explanation

Given

Refer Problem 10.57

The Henry’s law constant for O2 dissolved in arterial blood is 0.037mol/Latm .

Atmospheric pressure is 0.35atm .

The concentration of O2 in blood is calculated by the formula,

C=KH×p (1)

Where,

  • KH is the Henry’s law constant.
  • p partial pressure of O2 .

Partial pressure of the O2 gas is calculated by the formula,

p=m×P

Where,

  • m is the mole fraction of oxygen.
  • P is the atmospheric pressure.

The mole fraction of oxygen is 0.209.

Substitute the value of mole fraction and atmospheric pressure in above formula.

p=0.209×0.35atm=0.0731atm

Partial pressure of O2 gas is 0.0731atm .

Substitute the value of partial pressure and Henry’s law constant in equation (1).

C=0.037mol/Latm×0.0731atm=0.0027mol/L_

Therefore, the solubility of O2 in the blood of a climber on Mt. Everest is 0.0027mol/L_ .

(b)

Interpretation Introduction

To determine: The solubility of O2 in the blood of a scuba diver at 100 feet.

(b)

Expert Solution
Check Mark

Answer to Problem 10.59QP

Solution

The solubility of O2 in the blood of a scuba diver at 100 feet is 0.0231mol/L_ .

Explanation of Solution

Explanation

Refer Problem 10.57

The Henry’s law constant for O2 dissolved in arterial blood is 0.037mol/Latm .

Total pressure is 0.35atm .

The concentration of O2 in blood is calculated by the formula,

C=KH×p (1)

Where,

  • KH is the Henry’s law constant.
  • p partial pressure of O2 .

Partial pressure of the O2 gas is calculated by the formula,

p=m×P

Where,

  • m is the mole fraction of oxygen.
  • P is the atmospheric pressure.

The mole fraction of oxygen is 0.209.

Substitute the value of mole fraction and total pressure in above formula.

p=0.209×3atm=0.627atm

Partial pressure of O2 gas is 0.627atm .

Substitute the value of partial pressure and Henry’s law constant in equation (1).

C=0.037mol/Latm×0.627atm=0.0231mol/L_

Therefore, the solubility of O2 in the blood of a scuba diver at 100 feet is 0.0231mol/L_ .

Conclusion

  1. a. The solubility of O2 in the blood of a climber on Mt. Everest is 0.0027mol/L_ .
  2. b. The solubility of O2 in the blood of a scuba diver at 100 feet is 0.0231mol/L_ .

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Chapter 10 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 10 - Prob. 10.4VPCh. 10 - Prob. 10.5VPCh. 10 - Prob. 10.6VPCh. 10 - Prob. 10.7VPCh. 10 - Prob. 10.8VPCh. 10 - Prob. 10.9QPCh. 10 - Prob. 10.10QPCh. 10 - Prob. 10.11QPCh. 10 - Prob. 10.12QPCh. 10 - Prob. 10.13QPCh. 10 - Prob. 10.14QPCh. 10 - Prob. 10.15QPCh. 10 - Prob. 10.16QPCh. 10 - Prob. 10.17QPCh. 10 - Prob. 10.18QPCh. 10 - Prob. 10.19QPCh. 10 - Prob. 10.20QPCh. 10 - Prob. 10.21QPCh. 10 - Prob. 10.22QPCh. 10 - Prob. 10.23QPCh. 10 - Prob. 10.24QPCh. 10 - Prob. 10.25QPCh. 10 - Prob. 10.26QPCh. 10 - Prob. 10.27QPCh. 10 - Prob. 10.28QPCh. 10 - Prob. 10.29QPCh. 10 - Prob. 10.30QPCh. 10 - Prob. 10.31QPCh. 10 - Prob. 10.32QPCh. 10 - Prob. 10.33QPCh. 10 - Prob. 10.34QPCh. 10 - Prob. 10.35QPCh. 10 - Prob. 10.36QPCh. 10 - Prob. 10.37QPCh. 10 - Prob. 10.38QPCh. 10 - Prob. 10.39QPCh. 10 - Prob. 10.40QPCh. 10 - Prob. 10.41QPCh. 10 - Prob. 10.42QPCh. 10 - Prob. 10.43QPCh. 10 - Prob. 10.44QPCh. 10 - Prob. 10.45QPCh. 10 - Prob. 10.46QPCh. 10 - Prob. 10.47QPCh. 10 - Prob. 10.48QPCh. 10 - Prob. 10.49QPCh. 10 - Prob. 10.50QPCh. 10 - Prob. 10.51QPCh. 10 - Prob. 10.52QPCh. 10 - Prob. 10.53QPCh. 10 - Prob. 10.54QPCh. 10 - Prob. 10.55QPCh. 10 - Prob. 10.56QPCh. 10 - Prob. 10.57QPCh. 10 - Prob. 10.58QPCh. 10 - Prob. 10.59QPCh. 10 - Prob. 10.60QPCh. 10 - Prob. 10.61QPCh. 10 - Prob. 10.62QPCh. 10 - Prob. 10.63QPCh. 10 - Prob. 10.64QPCh. 10 - Prob. 10.65QPCh. 10 - Prob. 10.66QPCh. 10 - Prob. 10.67QPCh. 10 - Prob. 10.68QPCh. 10 - Prob. 10.69QPCh. 10 - Prob. 10.70QPCh. 10 - Prob. 10.71QPCh. 10 - Prob. 10.72QPCh. 10 - Prob. 10.73QPCh. 10 - Prob. 10.74QPCh. 10 - Prob. 10.75QPCh. 10 - Prob. 10.76QPCh. 10 - Prob. 10.77QPCh. 10 - Prob. 10.78QPCh. 10 - Prob. 10.79QPCh. 10 - Prob. 10.80QPCh. 10 - Prob. 10.81QPCh. 10 - Prob. 10.82QPCh. 10 - Prob. 10.83QPCh. 10 - Prob. 10.84QPCh. 10 - Prob. 10.85QPCh. 10 - Prob. 10.86QPCh. 10 - Prob. 10.87QPCh. 10 - Prob. 10.88QPCh. 10 - Prob. 10.89QPCh. 10 - Prob. 10.90QPCh. 10 - Prob. 10.91QPCh. 10 - Prob. 10.92QPCh. 10 - Prob. 10.93QPCh. 10 - Prob. 10.94QPCh. 10 - Prob. 10.95QPCh. 10 - Prob. 10.96QPCh. 10 - Prob. 10.97QPCh. 10 - Prob. 10.98QPCh. 10 - Prob. 10.99QPCh. 10 - Prob. 10.100QPCh. 10 - Prob. 10.101QPCh. 10 - Prob. 10.102QPCh. 10 - Prob. 10.103QPCh. 10 - Prob. 10.104QPCh. 10 - Prob. 10.105APCh. 10 - Prob. 10.106APCh. 10 - Prob. 10.107APCh. 10 - Prob. 10.108APCh. 10 - Prob. 10.109APCh. 10 - Prob. 10.111APCh. 10 - Prob. 10.110APCh. 10 - Prob. 10.112APCh. 10 - Prob. 10.113APCh. 10 - Prob. 10.114APCh. 10 - Prob. 10.115APCh. 10 - Prob. 10.116APCh. 10 - Prob. 10.117APCh. 10 - Prob. 10.118APCh. 10 - Prob. 10.119AP
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