`1.0` gm of moist sample of mixture of potassium chlorate `(KClO_(3))` and potassium chloride (KCl) was dissolved in water and solution was made upto 250 ml. This solution was treated with `SO_(2)` to reduce all `ClO_(3)^(-) " to " Cl^(-)` and excess of `SO_(2)` was removed by boiling . The total chloride was precipitated as silver chloride . The weight of precipitate was found to be `0.1435` gm. In another experiment , 25 ml of original solutions was heatod with 30 ml `0.2 N FeSO_(4)` and unused `FeSO_(4)` required `37.5 ` ml of `0.08 N KMnO_(4)` solutions .
Calculate the molar ratio of the `ClO_(3)^(-)` to the `Cl^(-)` in the given mixture
Given that , `ClO_(3)^(-) + 6l^(-) e^(2+) + 6H^(+) to Cl^(-) + 6Fe^(3+) + 3H_(2)O`
`3SO_(2) + ClO_(3)^(-) + 3H_(2)O to Cl^(-) + 3SO_(4)^(2-) + 6H^(+)`

Explanation

`ClO_(3)^(-)` is reduced to `Cl^(-) " by " SO_(2) " and " ClO_(3)^(-)` is also reduced to `Cl^(-) " by " Fe^(2+) ` , hence AgCl is formed due to total `Cl^(-)`
Meq. of `Fe^(2+) ` initially taken ` = 30 xx 0.2 = 6`
Meq. of `Fe^(2+)` unused ` = 37 . 5 xx 0.08 = 3`
` :. ` Meq of `Fe^(2+) " used " = 6.0 - 3.0 = 3.0`
Thus, Meq. of `ClO_(3)^(-) " in " 25 ml = 3.0`
Moles of `ClO_(3)^(-) " in " 25 ml = (3.0)/(1000 xx 6) = 0.0005`
`.^(+5)ClO_(3)^(-) to .^(-1)Cl^(-) ` (n-factor 6)
0.N `5 .... -1`
Thus , moles of `ClO_(3)^(-)` in 25 ml solution ` = 0.0005`
`ClO_(3)^(-)` is also reduced to `Cl^(-)` by `SO_(2)` in first experiment and precipitated as AgCl.
Thus , `Cl^(-)` formed from `ClO_(3)^(-) = AgCl` from `ClO_(3)^(-) = 0.0005`
Total AgCl formed both from actual and `Cl^(-) " from " ClO_(3)^(-) = 0.1435` gm
` = (0.1435)/(143.5) = 0.0010` mol
Thus, AgCl formed due to `Cl^(-) ` only = `0.0010 - 0.0005 = 0.0005` mol
Thus , `ClO_(3)^(-) " and " Cl^(-)` are in molar ratio ` = 1 : 1 `.