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The solubility of N 2 in water at 300   K and 500   torr partial pressure is 0 .01 g L − 1 . The solubility (in g L − 1 ) at 750   torr partial pressure is:

Options

  1. A0.0075
  2. B0.005
  3. C0 . 02
  4. D0.015

Correct answer

D. 0.015

Step-by-step solution

p   =  K H   X where p → Partial pressure of gas X 2 → Mole fraction of gas in solution K H  = Henry's law constant So, p ∝ solubility p 1 p 2  =   s 1 s 2   ⇒   500 0 . 01  =   750 x ∴   x  =  0 . 015   g / L

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