# Question #fe55f

##### 1 Answer

I have no idea what your Henry's law constant units are...

[It's possible that you are using a mol fraction version of Henry's law, in which

I get

**REGARDING THE HENRY'S LAW CONSTANT**

I cannot trust the Henry's law constant given here. There are no units given.

Using

#k_H ("mine")#

#= 6.1 xx 10^(-4) "mols solute" cdot cancel("L"^(-1) "soln") cdot "atm"^(-1) xx cancel"1 L soln"/(1000 cancel"g water") xx (18.015 cancel"g water")/"1 mol soln"#

#= 1.0989 xx 10^(-5)# #"atm"^(-1)#

Then in your case, you may have

#1/(k_(H)("mine")) = k_(H) ("yours") = "90998.8 atm"# , which is somewhat close to#100000# , I suppose...

**MOL FRACTION VERSION OF HENRY'S LAW**

So, I simply assume the version of **Henry's Law** shown below:

#P_(gas) = chi_(gas(l))k_H# where:

#chi_(gas(l)) = (n_(gas))/(n_(gas) + n_(solvent))# is themol fractionof the gas in the solution.#k_H# is theHenry's law constant(for the PARTICULAR gas at a SPECIFIED temperature in the SPECIFIED solvent!!). This WILL vary depending on your Henry's law equation, and there ARE multiple versions of this equation!! Keep the units straight.#P_(gas)# is thevapor pressureof the gas above the solution. Here, it is in#"atm"# .

**RELATING TO PARTIAL PRESSURE IN AIR**

Since the air vapor pressure is

#chi_(N_2(v)) = 0.78# .

So, the partial pressure of

#P_(N_2) = chi_(N_2(v))P_(t ot)#

#= 0.78 ("10 atm") = "7.8 atm"#

**GETTING MOLS OF N2**

Thus, using the Henry's law constant of *in* solution:

#chi_(N_2(l)) = P_(N_2)/k_H = ("7.8 atm")/("90998.8 atm")#

#=# #0.00008572#

Since we have

#"10 L water" ~~ "10000 g water" ~~ "555.09 mols water"#

And so, the mol fraction allows us to solve for the mols of gas in here:

#chi_(N_2(l)) = 0.00008572 = n_(N_2)/(n_(N_2) + n_(H_2O))#

#= n_(N_2)/(n_(N_2) + 555.09)#

#=> 0.00008572n_(N_2) + 0.00008572(555.09) = n_(N_2)#

#=> color(blue)(n_(N_2)) = (0.04758)/(1 - 0.00008572) = 0.0476#

#~~# #color(blue)("0.048 mols N"_2 " in the solution")#