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I'm studying for my chemistry final and having a hard time with this question on one of the practice exams. Here's the question:

At $20$ degreees Celsius, Henry's Law constant for Ar is half of that for Kr. If a gaseous Ar/Kr mixture containing 20% molar fraction of Kr is placed over water at $20$ degrees Celcius, what is the ratio of concentrations of Ar to Kr in water?

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  • $\begingroup$ Since there is a 20% molar fraction of Kr, that means the mixture contains 4 parts Ar to 1 part Kr. 4 parts * 1/2 gas constant = 2 times more Ar than Kr. $\endgroup$ – Neurax May 2 '13 at 22:59
  • $\begingroup$ this would make a good answer if you expand on it. $\endgroup$ – Ben Norris May 2 '13 at 23:25
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Henry's Law is mathematically expressed by $$p=k_Hc$$ where $p$ is the partial pressure above the liquid, $k_H$ is the Henry constant for the specific gas at the given temperature, and $c$ is the concentration of gas in liquid. If you set up one equation for each gas and divide the two, you get $$\frac{p_{Ar}}{p_{Kr}}=\frac{k_{H, Ar}}{k_{H, Kr}}\frac{c_{Ar}}{c_{Kr}}.$$ You are given both the ratio of partial pressures and the ratio of the Henry's Law constants; the ration of concentrations should be easily calculated from that point.

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