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The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is
- a)4.0 x 10 − 4
- b)4.0 x 10 − 5
- c)5.0 x 10 − 4
- d)4.0 x 10 − 6
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The Henry's law constant for the solubility of N2 gas in water at ...
To find the number of moles of N2 dissolved in water, we can use Henry's Law equation:
n(N2) = K(H) * P(N2)
where n(N2) is the number of moles of N2 dissolved in water, K(H) is the Henry's Law constant, and P(N2) is the partial pressure of N2.
Given:
Henry's Law constant (K(H)) = 1.0 x 10^5 atm
Partial pressure of N2 (P(N2)) = 0.8 * 5 atm = 4 atm
Substituting these values into the equation, we have:
n(N2) = (1.0 x 10^5 atm) * (4 atm)
= 4.0 x 10^5 mol
So, the number of moles of N2 dissolved in 10 moles of water is 4.0 x 10^5 mol.
n(N2) = K(H) * P(N2)
where n(N2) is the number of moles of N2 dissolved in water, K(H) is the Henry's Law constant, and P(N2) is the partial pressure of N2.
Given:
Henry's Law constant (K(H)) = 1.0 x 10^5 atm
Partial pressure of N2 (P(N2)) = 0.8 * 5 atm = 4 atm
Substituting these values into the equation, we have:
n(N2) = (1.0 x 10^5 atm) * (4 atm)
= 4.0 x 10^5 mol
So, the number of moles of N2 dissolved in 10 moles of water is 4.0 x 10^5 mol.
Community Answer
The Henry's law constant for the solubility of N2 gas in water at ...
Hiii Vivek .. It would be something like.. 1. use Henrys law constant i.e Partial pressure=K*x
2. Then use Ptotal=partial pressure ×mole fraction and find the no of moles from here... Regards
2. Then use Ptotal=partial pressure ×mole fraction and find the no of moles from here... Regards
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The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer?
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The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer?.
The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer?.
Solutions for The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure isa)4.0 x 10 − 4b)4.0 x 10 − 5c)5.0 x 10 − 4d)4.0 x 10 − 6Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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