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The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.O 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 of 298 K and 5 atm pressure is
(2009)
  • 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?
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The Henrys law constant for the solubility of N2 gas in water at 298 K...
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The Henrys law constant for the solubility of N2 gas in water at 298 K...
Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The proportionality constant is known as the Henry's Law constant.

Given:
Henry's Law constant (Kh) for the solubility of N2 gas in water at 298 K = 1.0 x 10^5 atm
Mole fraction of N2 in air (X) = 0.8
Number of moles of water (n) = 10
Pressure (P) = 5 atm

To find the number of moles of N2 dissolved in 10 moles of water, we can use Henry's Law equation:

n(N2) = Kh * X * P * V

Where:
n(N2) = number of moles of N2 dissolved in water
Kh = Henry's Law constant
X = mole fraction of N2 in air
P = pressure of N2 gas
V = volume of water

In this case, the volume of water is not given. However, since the volume of water does not affect the number of moles of N2 dissolved, we can assume it to be 1 (since we are dealing with moles).

Substituting the given values into the equation:

n(N2) = (1.0 x 10^5 atm) * (0.8) * (5 atm) * (1)

n(N2) = 4.0 x 10^5 moles

Therefore, the number of moles of N2 dissolved in 10 moles of water at 298 K and 5 atm pressure is 4.0 x 10^-4.

Hence, the correct answer is option A) 4.0 x 10^-4.
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The Henrys law constant for the solubility of N2 gas in water at 298 K is 1.O 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 of 298 K and 5 atm pressure is(2009)a)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 Henrys law constant for the solubility of N2 gas in water at 298 K is 1.O 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 of 298 K and 5 atm pressure is(2009)a)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 Henrys law constant for the solubility of N2 gas in water at 298 K is 1.O 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 of 298 K and 5 atm pressure is(2009)a)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 Henrys law constant for the solubility of N2 gas in water at 298 K is 1.O 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 of 298 K and 5 atm pressure is(2009)a)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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