The sulphur content of a steel sample is determined by converting it to `H_(2)S` gas, absorbing the `H_(2)S` in 10 mL of 0.005 M `I_(2)` and then back titrating the excess `I_(2)` with 0.002 M `Na_(2)S_(2)O_(3)`. If 10 mL `Na_(2) S_(2)O_(3)` is required for the titration , how many milligrams of sulphur are contained in the sample ?
Reactions : `H_(2)S +I_(2) rarr S + 2I^(-)+2H^(+) " " , " " I_(2)+2S_(2)O_(3)^(2-) rarr 2I^(-)+S_(4)O_(6)^(2-)`

Explanation

Used millimoles of `I_(2)` = (m.moles of `I_(2)` taken initially)-`("m.moles of hypo used")/(2)`
`=0.005xx10-0.002xx(10)/(2)`
`=0.04 = "millimoles of " H_(2)S`
`:. ` Weight of sulphur `=0.04xx10^(-3)xx32xx10^(3)mg=1.28 mg`.