`1 g` sample of `AgNO_(3)` is dissolved in `50 mL` of water, It is titrated with `50 mL` of `KI` solution. The `Agl`percipitated is filtered off. Excess of `KI` filtrate is titrated with `M//10KIO_(3)` in presence of `6 M HCl` till all `I^(-)` converted into `ICI`. It requires `50 mL` of `M//10 KIO_(3)` solution. `20 mL` of the same stock solution of `KI` requires `30 mL` of `M//10KIO_(3)` under similar conditions. Calculate `%` of `AgNO_(3)` in sample. The reaction is
`KIO_(3)+2KI+6HClrarr3ICl+3KCl+3H_(2)O`

Explanation

`KIO_3+2KI+6"HCl"to3ICI+KCl+3KCl+3H_2O`
30 " mL of " `(M)/(10)KIO_3=3 m" mol of "KIO_3`
`=6 m" mol of "KI`
`20 " mL of " KI` contains 6 m mol
`50 mL` of `KI` contains`=(6)/(20)xx50=15m" mol of "KI`
Total KI`15mol`
`50" mL of " Kio_3=5 m" mol of "KIO_3=10 m" mol of "KI`
`therefore` Excess of `KI=10mmol`
`KI` used up`=15-10=5mmol`
`AgNO_3+KItoAgI+KNO_3`
`5 m" mol of "KI=5 m" mol of "AgNO_3`
`=5xx10^(-3)xx170g of AgNO_3=0.85g`
`% of AgNO_3=85%`