`12. g` of an impure sample of arsenious oxide was dissolved in water containing `7.5 g` of sodium bicarbonate and the resulting solution was diluted to `250 mL`. `25 mL` of this solution was completely oxidised by `22.4 mL` of a solution of iodine. `25 mL` of this iodine solution reacted with same volume of a solution containing `24.8 g` of sodium thiosulphate `(Na_(2)S_(2)O_(3).5H_(2)O)` in one litre. Calculate teh percentage of arsenious oxide in the sample ( Atomic mass of `As=57`)

Explanation

`As_(2)O_(3)` sample`=12.0 g` , it reacts with `NaHCO_(3)` to give `Na_(3)AsO_(3)`. Its reaction with `I_(2)` shows the changes:
`{:(As_(2)^(3+),rarr,As_(2)^(5+)+,4e, ,),(I_(2)+2e,rarr,2I^(-),,):}`
Meq.of `As_(2)O_(3)` in `25 mL=` Meq.of `I_(2)`
`=22.4xxN`....(1)
Also `N` of `I_(2)` can be evaluated as:
The reaction are: `I_(2)+2erarr2I^(-)`
`2S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+2e`
Meq.of `I_(2)=` Meq.of hypo
`=NxxV`
`Nxx25=(24.8)/(248xx1)xx25`
`therefore N_(12)=(N)/(10)`
`therefore "Meq.of" As_(2)O_(3)"in" 25 mL=22.4xx(1)/(10)=2.24`
or `"Meq.of" As_(2)O_(3)"in" 250 mL=2.24xx(250)/(25)=22.4`
or `(w)/(E.wt.)xx1000=22.4`
`(w)/((198)/(4))xx1000=22.4`
`therefore w_(As_(2)O_(3))=(22.4xx198)/(4xx1000)=1.1088`
`therefore % "of" As_(2)O_(3)=(1.1088)/(12)xx100=9.24%`