`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=74`)

Explanation

`As_(2)O_(3)=12.0g`. It reacts with `NaHCO_(3)` to give `Na_(3)AsO_(3)`. Its reaction with `I_(2)` shows the changes.
(a). `undersetunderset(2x=2)(2x-6=0)(overset(+3)(As_(2))O_(3))toundersetunderset(2x=10)(2x-10=0)(overset(+5)(As_(2))O_(5))+4e^(-)(n=4)`
(b). `I_(2)+2e^(-)to2I^(ɵ)(n=2)`
(c). `2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2e^(-)` (n=(2)/(2)=1)`
`m" Eq of "As_(2)O_(3)` in `25mL=m" Eq of "I_(2)=22.4xxN`
Aso, `m" Eq of "I_(2)=m" Eq of "hypo (S_(2)O_(3)^(2-))`
`=NxxV`
`Nxx25=(24.8)/((248)/(1))xx25`
`[Ew of Na_(2)S_(2)O_(3).5H_(2)O=(248)/(1)(n=1)]`
`therefore_(I_(2))=(1)/(10)`
`m" Eq of "As_(2)O_(3)` in `25 m22.4xx(1)/(10)=2.24`
or `m" Eq of "As_(2)O_(3)` in `250mL=(2.24xx250)/(25)=22.4`
`(W)/(Ew)xx10^(3)=22.4`
`W_(As_(2)O_(3))=(22.4xx198)/(4xx10^(3))=1.1088`
`% of As_(2)O_(3))=(22.4xx198)/(4xx10^(3))=1.1088`
`% of As_(2)O_(3)=(1.1088)/(12)xx100=9.24%`