The Henry's law constants for N2 and O2 in water are 8.6E-4 /kPa and 4.4E-4 /kPa respectively There are approximately 1E5 Pa in 1 atm. Air is approximately 79% nitrogen and 21% oxygen. What is the approximate ratio of nitrogen to oxygen molecules dissolved in a glass of water that has been left exposed to 1 atm of air pressure for a long time?
Thanks for your help
Hmm.. well first off I'm pretty sure you have the units wrong on the constants.. they should be kPa/M that way you end up with kPa after multiplying by concentration.
So henry's law: partial pressure = constant*concentration and we want to solve for concentration so: concentration = pressure/constant
Now let's work in kPa because of the constant's units. 1 atm has a partial pressure of .79 N2 and .21 O2 or 7900 Pa (7.9 kPa) and 2100 Pa (2.1 kPa), respectively.
Concentration of N2 = 7.9/8.6*10^-4 = (approx) 9200 M
Concentration of O2 = 2.1/4.4*10^-4 = (approx) 4700 M
Numbers don't seem very right so I prob made a mistake but my answer would be about 2:1.
Post the answer so I can see where (if) i went wrong and help you out.