An impure sample of `As_(2)O_(3)` weighitng `20 gm` was dissolved in water containing `10 gm` of `NaHCO_(3)` and the final solution was diluted to `400ml`.`20ml` of this solution was oxidised by `25ml` of a solution of `l_(2)`. Another solution of hypo containing `1.24 gm` of `Na_(2)S_(2)O_(3).5H_(2)O` in `40ml` was used to exactly reduce `25 ml` of same `l_(2)` solution (impurities inert) calculate `%` fo `AS_(2)O_(3)` in the sample (at. wt. of `As=75`)

Answer

`00049.50`

Explanation

`00049.50`
`M_("hypo")=(1.24xx1000)/(248xx40)=0.125`
meq of hypo`=` meq of `I_(2)` in `25 ml =0.125xx1xx40=5`
meq of `l_(2)=` meq of `As_(2)O_(3)=5`
meq of `AS_(2)O_(3)` in `20 ml` solution `=5`
so meq of `AS_(2)O_(3)` in `400 ml=(5xx400)/20=100`
mole of `AS_(2)O_(3)=0.05xx198=9.9g`
wt. % `=9.9/20xx100=49.5%`