1 gm sample of AgNO_(3) is dissolved in 50 ml of water . It is titrated with 50 ml of Kl Solution : The Agl preciplitated is filtered off . Excess of Kl in filtrate is titrated with M/10 KlO_(3) in presence of 6 M HCl till all l^(-) converted into ICl. It requires50 ml of M/10 KlO_(3) Soluition : 20 ml of the same stock solution of Kl requires 30 ml of M/10 KlO_(3) under similar conditions. Calculate % of AgNO_(3) in the sample . The reaction is KlO_(2) + 2Kl + 6HCl to 3lCl + 3KCl + 3H_(2)O .

1 gm sample of AgNO_(3) is dissolved in 50 ml of water . It is titrate

1.	1 gm sample of AgNO_(3) is dissolved in 50 ml of water . It is titrated with 50 ml of Kl Solution : The Agl preciplitated is filtered off . Excess of Kl in filtrate is titrated with M/10 KlO_(3) in presence of 6 M HCl till all l^(-) converted into ICl. It requires50 ml of M/10 KlO_(3) Soluition : 20 ml of the same stock solution of Kl requires 30 ml of M/10 KlO_(3) under similar conditions. Calculate % of AgNO_(3) in the sample . The reaction is KlO_(2) + 2Kl + 6HCl to 3lCl + 3KCl + 3H_(2)O .
Answer» Solution :`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 Klin 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)) = 0.85gm= 85 %` `("Moleof " AgNO_(3))/("Mole of " K_(t)) = 1/l` ` :. ` If equivalent WEIGHT of `Kl = M/2` Equivalent weight of `AgNO_(3) = M/2`