Class 12 Exam  >  Class 12 Questions  >  Air contains O2 and N2 in the ratio of 1:4 . ... Start Learning for Free
Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively?
Most Upvoted Answer
Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of so...
Ratio of Solubility of O2 and N2 in Water


Given Information:


  • Air contains O2 and N2 in the ratio of 1:4

  • Henry's constant for O2 = 3.30×107 Torr

  • Henry's constant for N2 = 6.60×107 Torr



Solution:

We know that the mole fraction of O2 and N2 in air is 1:4.


Let the mole fraction of O2 and N2 dissolved in water be x and y respectively.


According to Henry's law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Therefore, we can write:


x = (Henry's constant for O2) x (Partial pressure of O2 in air)

y = (Henry's constant for N2) x (Partial pressure of N2 in air)


Let the partial pressure of O2 and N2 in air be P and 4P respectively (as they are in the ratio of 1:4).


Then, x = (3.30×107 Torr) x (P/T) and y = (6.60×107 Torr) x (4P/T), where T is the total pressure of air (which is atmospheric pressure).


Now, we can write the ratio of solubility of O2 and N2 as:


x:y = (3.30×107 Torr) x (P/T) : (6.60×107 Torr) x (4P/T)


x:y = 1:2


Conclusion:

The ratio of solubility of O2 and N2 in water is 1:2.
Community Answer
Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of so...
Explore Courses for Class 12 exam
Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively?
Question Description
Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively?.
Solutions for Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively? defined & explained in the simplest way possible. Besides giving the explanation of Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively?, a detailed solution for Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively? has been provided alongside types of Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively? theory, EduRev gives you an ample number of questions to practice Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of solubility in terms of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O2 and N2 are 3.30×10 raised to the power 7 and 6.60×10raised to the power 7 Torr respectively? tests, examples and also practice Class 12 tests.
Explore Courses for Class 12 exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev