Question #fe55f

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 #k_H# (in #"atm"#) is approximately the numerical reciprocal of the version in units of #"M/atm"#.]

I get #"0.048 mols N"_2(g)# in the solution, using #k_H = 90998.8# #"atm"#. You must adjust accordingly using your #100000# #"atm"#.

REGARDING THE HENRY'S LAW CONSTANT

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

Using #k_H# in #"M/atm"#, if I convert my Henry's law constant to what I think is yours...

#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 the mol fraction of the gas in the solution.
• #k_H# is the Henry'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 the vapor pressure of the gas above the solution. Here, it is in #"atm"#.

RELATING TO PARTIAL PRESSURE IN AIR

Since the air vapor pressure is #"10 atm"#, we know that by the definition of partial pressure, and knowing the percent of #"N"_2# in air is #78%#, the mol fraction of #"N"_2# in air is

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

So, the partial pressure of #"N"_2# in the air above the solution is

#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 #"N"_2# in water at #"298 K"# (not the temperature which was not given in the question!), which I converted above to be #90998.8# #"atm"#, I get a mol fraction of #"N"_2# in solution:

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

#=# #0.00008572#

Since we have #"10 L"# of water, assuming its density is #"1000 g/L"#, and the gas hardly changes the volume of the water:

#"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")#