The solubility of Ar in water at 25 °C is 1.6 ⋅ 10-3 M when the pressure of the Ar above the solution is 1.0 atm. The solubility of Ar at a pressure of 2.5 atm is M. The solubility of Ar in water at 25 °C is 1.6 10-3 M when the pressure of the Ar above the solution is 1.0 atm. The solubility of Ar at a pressure of 2.5 atm is M. 1.6 ⋅ 10-3 1.6 ⋅ 103 4.0 ⋅ 10-3 6.4 ⋅ 10-4 7.5 ⋅ 10-2

SAr = 3.75 E-3 M ≅ 4.0 E-3 M

Explanation:

Henry's law:

Sg = (Kh)*(Pg)

∴ Sg: gas solubility (mol/L)

∴ Kh: Henry's constant ⇒ Kh Ar = 1.5 E-3 mol/atm.Lfrom literature

∴ Pg: Partial pressure of gas on solution (atm)

If Pg = 2.5 atm

⇒ Sg = (1.5 E-3 mol/atm.L)(2.5 atm)

⇒ Sg = 3.75 E-3 M

I’m not sure what your trying to say here.. can you clarify again?

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