gen chem question from datqvault

[Menthol]

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The solubility of dissolved oxygen in water is 1.25 x 10-4 M at 25°C, where the mole fraction of oxygen is 0.21 and atmospheric pressure is 1.0 atm. In a pure oxygen atmosphere at the same pressure, what would the solubility be?

how can i solve this problem without knowing a total pressure and a pure oxygen atmosphere value ?
thank you

 

[cj7]

this is a proportion
(1.25 x 10-4 M)/.21atm = (new solubility)/1atm
new solubility= (1.25 x 10-4 M)/.21atm = 40.4M

 

[dent4]

how is that a proportion if it says the mole fraction is .21 not .21atm

 

[Menthol]

how is that a proportion if it says the mole fraction is .21 not .21atm

Yeah if the partial pressure was given, then that formula would be possible
but without it, i have no idea !stresseddd

 

[CedarZ4]

The solubility of dissolved oxygen in water is 1.25 x 10-4 M at 25°C, where the mole fraction of oxygen is 0.21 and atmospheric pressure is 1.0 atm. In a pure oxygen atmosphere at the same pressure, what would the solubility be?

how can i solve this problem without knowing a total pressure and a pure oxygen atmosphere value ?
thank you

Simply, Henry's Law.

Henry's Law states that the amount of dissolved gas in a liquid is proportional to the partial pressure of the gas.

As you have mentioned, you aren't given partial pressure. However, recall that partial pressure is also proportional to the mole ratio.

So, they tell you the mole fraction is .21 (or .21 mole O2/Whole).

Therefore, Dissolved Gas is Proportional to Partial Pressure which is proportional to Concentration (Mole Fraction) therefore Dissolved Gas is Proportional to Concentration.

Old Dissolved/New Dissolved = Old Mole Ratio/New Mole Ratio.
1.25 x10^-4 / X = .21 / 1 (because in pure oxygen atmosphere, mole ratio is 1/1)

Hope this helps.