[arlibera]

"at 20 degrees Celsius, the equilibrium partial pressure of SO2 in gas over a solution of 0.001403 mole fraction SO2 in water is 0.0342 atm. Estimate the concentration (mole fraction) of SO2 in rain water if there is 0.0012 mol fraction SO2 in air and the air pressure is 1 atm."

*First I found the total equilibrium pressure. I divided .0342 atm by the mole fraction .001403 to get 24.37 atmospheres/2.5e6 pascals.
*I plugged this into the ideal gas equation (Vm=RT/P) to get the equilibrium molar concentration for the solution which is .00099mol/m3
*I then found the Vm for the non-equilibrium state, using the equation Vm=RT/P
*Vm came out to be .0245 mol/m3
* I used this to calculate the concentration of the precipitate by subtracting the two molar volumes.
Vm1-Vm2 = .0235mol/m3
*Finally, I multiplied this number by the mol fraction of SO2 to find the concentration precipitated.
=0.0000282mol/m3

Can anyone tell me if I solved this problem correctly?

[Borek]

I divided .0342 atm by the mole fraction .001403 to get 24.37 atmospheres/2.5e6 pascals.

Pressure is for a gaseous phase, mole fraction is for a liquid phase. You are dividing apples by oranges.

Use Henry's law.

[curiouscat]

=0.0000282mol/m3

Not at all relevant to your question, but it's better practice to resort to Scientific Number convention when you have that many decimals.

i.e. 2.82×10^-5

Someday it'll save you from an arithmetic error.