A sample of pure aniline was dissolved in HCl and diluted to 100 mL with `H_2O`. 20 " mL of " liquid was treated with 25 " mL of " 0.017 M `KBrO_3` and about 10 g KBr was added to form `Br_2`. After 10 min, an excess of KI was added and the liberated `I_2` was titrated with 12.92 " mL of " 0.12 M `Na_2S_2O_3`. Calculate the weight of aniline taken.

Explanation

`BrO_3^(ɵ)+5Br^(ɵ)+6H^(o+)to3Br_2+H_2O`
`1 mol` of `BrO_3^(ɵ)-=3 mol` of `Br_2`
`25xx0.017 mmol =3xx25xx0.017 mmol` of `Br_2`
Reactions involved are
`2I^(ɵ)+Br_2toI_2+2Br_(ɵ)`
`I_2+2e^(-)to2I^(ɵ)(x=2)`
`2S_2O_3^(2-)toS_4O_6^(2-)+2e^(-)(x=(2)/(2)=1)`
Excess of `Br_2-=` moles of `I_2`
m" Eq of "`Na_2S_2O_3=12.92xx0.12xx1`
`=12.92xx0.12 mEq` of `I_2`
`=(12.92xx0.12)/(2) mmol` of `I_2`
`Br_2` reacted with aniline `=(3xx25xx0.017-(12.92xx0.12)/(2))`
`=12.75-0.7725`
`=0.4998 mmol` of `Br_2`
`approx 0.5 mmol` of `Br_2`
3 m" mol of "`Br_2=1` m" mol of "aniline
0.5 m" mol of "`Br_2=(1)/(3)xx0.5` m" mol of "aniline
`=(0.5)/(3)xx10^(-3)xx93=0.0155g`