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Calculate the solubility of certain gas in solvent with pressure 3 atm at 25^ C (Henry's law constant is 3.0 10⁻² , mol dm ⁻³ , atm ⁻¹ )

Options

  1. A0.07 M
  2. B0.08 M
  3. C0.09 M
  4. D0.1 M

Correct answer

C. 0.09 M

Step-by-step solution

According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. S = K_H P Given: K_H = 3.0 10⁻² mol dm ⁻³ atm ⁻¹ P = 3 atm Substituting the values: S = (3.0 10⁻² mol dm ⁻³ atm ⁻¹) 3 atm S = 9.0 10⁻² mol dm ⁻³ Since 1 mol dm ⁻³ = 1 M , the solubility is 0.09 M . Answer: 0.09 M

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