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`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`) |
Answer» `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%` |
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