1.0 g of `AgNO_3` is dissolved in 50 " mL of " water It is titrated with 50 " mL of " KI solution. The Agl precipitated is filtered off. Excess of KI in the filtrate is titrated with `(M)/(10)` `KIO_3` in the presence of `6MHCl` till all `I^(ɵ)` converted into ICI. It requires 50 " mL of " `(M)/(10)` `KIO_3` solution. 20 " mL of " the same solution of KI requires 30 " mL of " `(M)/(10)KIO_3` under the same conditions. Determine the percentage of `AgNO_3` in the sample.

Explanation

`AgNO_(3) + Kl to Agl downarrow + KNO_(3)`
Ag present in `AgNO_(3)` is removed as Agl by adding 50 ml of Kl solution of which 20 ml requires 30 ml of M/10 `KlO_(3)`
Meq. of Kl in 20 ml = Meq. of `KlO_(3) = 30 xx 1/10 xx 4`
(For `KlO_(3) , l^(5+) + 4e^(-) to l^(+) l_(-) to l^(2+) + 2e^(-) ` )
`:." Wq, wt. of " Kl = M/2 `
`:. ` Meq. of Kl in 50 ml added to `AgNO_(3) = (30 xx 4 xx 50)/(10 xx 20) = 30 `
Meq. of Kl left unused by `AgNO_(3) = 50 xx 1/10 xx 4 = 20`
` :. ` Meq. of Kl used for `AgNO_(3) = 30 - 20 = 10`
` :. ` Meq. of `AgNO_(3) = 10`
`W_(AgNO_(3))/(170) xx 2 xx 1000 = 10`
`:. W_(AgNO_(3)) = 0.85 gm = 85 %`
`("Mole of " AgNO_(3))/("Mole of " K_(t)) = 1/l`
` :. ` If equivalent weight of `Kl = M/2`
Equivalent weight of `AgNO_(3) = M/2`