[ineedhelp10]

Hi i have already asked for help on this about a week ago and didn't get anywhere.  so I'm going to start smaller.

The question I'm answering:

"look up Henry's law constant and use it to calculate the amount of so2 gas which would dissolve in 500ml of pure h20 placed near and active volcano.  you may assume equilibrium conditions, and ignore any reactions with h20.  the atmosphere surrounding the volcano contains about 5% so2.  convert your answer to grams."

I'm using the equation [x, solvant] = K_h.P(X_(g))

so i found the constant to be 0.81 atm-L/mole
and now i need to work out the P(_(g))

i have been reading and the partial pressures of a mixture of gasses = the total pressure. so I've been told that 5% of the atmosphere is so2.  thus in 100mols of the atmosphere 5 mols would be so2.  

therefore would the following be correct

SO₂ = 0.05x0.81 which = 0.0405 mol.L^-1 (this is for 1 litre of h20)

0.0405/2 = 0.02025 mol per 500ml

to get grams from this i then worked out how much 1 mol of so2 weighs (s=16 o=8x2)  so 32g's per mol.  
32/10000*2025 = 6.49g  

is this correct or have i gone seriously wrong.  thanks for any input in advance

[MrTeo]

so i found the constant to be 0.81 atm-L/mole
and now i need to work out the P(_(g))

Almost correct... just one thing: if k_h is 0.81 L∙atm∙mol^-1 you can easily see that the units won't work with the formula you've written (c_gas=k_h∙p_gas) and you'll have to change something to make them fit in your equation... (Henry's law can be written in many different ways, just changing the value of the constant)