A mixture of two gases, `H_(2)S` and `SO_(2)` is passes throgh three beakers successivly. The first beaker contains `Pb^(2+)` ions, which absobs all `H_(2)S` to form `PbS`. The second beaker contains `25mL` of `0.0396 N I_(2)`. Which ocidises all `SO_(2)` to `SO_(4)^(2-)` . The third beaker contains `10 mL` of `0.0345 N` thisulphate solution ti retain any `I_(2)` carried over from the second absorber. The solution from first absorber was made acidic and treated with `20 mL` of `0.0066 M K_(2) Cr_(2) O_(7)`, acidic and treated with `20mL` of `0.006M K_(2) Cr_(2)O_(7)` which converted `S^(2-)` to `SO_(2)`. The excess dichromate was reacted with solid `KI` and the liberated iodine required `7.45mL` of `0.0345N Na_(2)S_(2)O_(3)` solution. The solution in the second and thrid absorbers were combined and the resulatant iodide was treated with `2.44 mL` fo the same solution of thisulphate. Calculate the conventrations fo `SO_(2)` and `H_(2)S` in `(mg)/"litre"` of the sample.

Answer

conc of `SO_(2) = 0.72mg//"litre"`
conc. Of `H_(2)S = 0.125 mg.."litre"`