Solubility in different pressure conditions

I am working on my scholarship exam practice. I believe this exam assumes high school + first year university knowledge. And I'm not quite sure what I did wrong. Could you please have a look?



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> The solubility of oxygen in 1.0 L water is 28 mL at 25 °C and 1.0 atm.
>  How much oxygen can be dissolved in 1.0 L of water at 25 °C and 4.0 atm?
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The answer remains 28 mL but IMO I thought when pressure goes up 4 times while other factors are constant, the solubility would do that as well. Please help on where I missed and what topics I should focus on specifically.




As noted by Wikipedia, 



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> "[Henry's Law](https://en.wikipedia.org/wiki/Henry%27s_law) is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid."
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So you are absolutely right. 4 times the pressure of oxygen should result in 4 times as much oxygen being dissolved. 


The answer key for the problem must be wrong.



> 
> User Delta\_G made an interesting observation/correction. The problem is asking about the volume of gas that dissolves, not the concentration of oxygen in the water. So 28 ml at 4 atm of oxygen gas contains four times as much oxygen as 28 ml at 1 atm of the gas. So in either case 28 ml of the gas phase dissolves into the water. 
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 5

This is a confusing question because, while solubilities **can** be reported in mL/L, there can be ambiguity when choosing a pressure during conversion to this unit, for instance using the following equation to convert from molarity $c$ to volume/volume units:


$$ \rho = \frac{cRT}{p}$$


[In this online data page](https://www.engineeringtoolbox.com/oxygen-solubility-water-d_841.html), for instance , in some columns the solubility is reported in mL/L, converted to this unit using throughout a pressure of $\pu{1 atm}$ (the source of the data cannot be verified), even as the partial pressure of oxygen $p\_{O\_2}$ is increased *above* $\pu{1 atm}$. 


In the OP the volume refers presumably to an equivalent volume of oxygen gas at the (partial) pressure of the gas above the liquid.


Using the numbers from the OP, assuming the gas is ideal, then 


$ c=\frac{101325\times28\times 10^{-6}}{8.3145\times298.15} \pu{M} =\pu{ 1.1 \times 10^{-3}M}$


when $p = \pu{1 atm}$.


On the other hand, if $p = \pu{4 atm}$


$ c=\frac{4\times101325\times28\times 10^{-6}}{8.3145\times298.15} \pu{M} =\pu{ 4.6\times 10^{-3} M}$


so the solubility is the same ($\pu{28 mL/L}$) if described in terms of volume at the given pressure, but $\times 4$ greater when regarded as a molar concentration.


Note by the way that according to [a number](https://water.usgs.gov/owq/FieldManual/Chapter6/table6.2_6.pdf) of [sources](http://www.colby.edu/chemistry/CH331/O2%20Solubility.html) the solubility at $\pu{25 ^\circ C}$ is $\pu{258 \mu M}$ (~$\pu{8.2 mm Hg}$) at $p\_{O\_2}=\pu{1 atm}$, and $\pu{1.0 mM}$ at $p\_{O\_2}=\pu{4 atm}$.


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