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The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 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 inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 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 inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 10–6Correct answer is option 'A'. Can you explain this answer?.

Solutions for The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 10–6Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 10–6Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 10–6Correct answer is option 'A'. Can you explain this answer?, a detailed solution for The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 10–6Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 10–6Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)a)4.0× 10– 4b)4.0 × 10–5c)5.0 × 10– 4d)4.0 × 10–6Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.