`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`

Answer

`(85%)`

Explanation

The reaction is
`KIO_(3) + 2KI + 6HCl rarr 3ICl + 3KCl + 3H_(2)O`
`KIO_(3)` required for `20 mL` original `KI` solution = 3 millimol.
`rArr 7.5` millimol `KIO_(3)` would be required for original `50 mL KI`.
`rArr` Original `50 mL KI` solution contain 15 millimol of `KI`. ltbr. After `AgNO_(3)` treatment 5 millimol of `KIO_(3)` is required, i.e., 10 millimol `KI` is remaining.
`rArr 5` millimol `KI` reacted with `5` millimol of `AgNO_(3)`.
`rArr` Mass of `AgNO_(3) = (5)/(1000) xx 170 = 0.85 g`
`rArr` Mass percentage of `AgNO_(3) = 85%`