Henrys law calculation question?

[David Johnson]

Please could you help with the following question:

For water at 20 degrees the constant Kb, in the Henry's Law equation (Pb = XbKb), is equal to 4 x 10^9 Pa for oxygen and 8.3 x 10^9 Pa for nitrogen. Assuming the composition of air is 20 mole % oxygen and 79 mole % nitrogen, show that at atmospheric pressure and 20 degrees the volumes of oxygen and nitrogen dissolved in 1dm3 of water 6.8 and 12.9 respectively.

Ive been trying this question for a long time and dont seem to be getting anywhere. A method of how to solve this probem will be very much appreciated.

Thanks

[azteccameron1]

Oh dear the graveyard shift agen..thort Henry still playd 4 barca!..
neways..Rauol or sum law says..the partial press of gas =concn of gas*tot press=0.79x10^5Pa for Nitrogen and wif ur dogy figs..0.2x10^5Pa for O2..
..an using ur (where u get these consts??)..for Henrys law..
We get concn(N)= 0.79x10^5/8.3x10^9=9.52x10^-6mols/mol water..
sim for concn(O)=0.2x10^5/4x10^9= 5x10^-6mols/mol water..
now 1mol water=18g=18 ml..
..So [N2]=9.52x10^-6*1000/18= 5.289x10^-4moles/litre dissolved and at temp of 20DegC = volume of 12.89cm^3
..for [O2]=5x10^-6*1000/18=2.778x10^-4 moles/litre..
> 6.68cm^3 dissolved`per litre..bet u forgot to correct Avogs law 4 temp!..ie its 22.4*293/273 for molvol.