One gram of commercial `AgNO_(3)` is dissolved in 50 mL of water. It is treated with 50 mL of a KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filtrate is titrated with `(M//10) KIO_(3)` solution in presence of 6 M HCl till all `I^(-)` ions are converted into ICl. It required 50 mL of `(M//10) KIO_(3)` solution. 20 mL of the same stock solution of KI requires 30 mL of `(M//10) KIO_(3)` under similar conditions. Calculate the percentage of `AgNO_(3)` in the sample.
Reaction : `KIO_(3) + 2KI + 6HCl rarr 3ICl + 3KCl + 3H_(2)O`

Explanation

Number of millimoles of `KIO_(3)` in 30 mL of solution = `"molarity" xx "volume in mL" = 1/10 xx 30 = 3`
Given reaction : `KIO_(3) + 2KI + 6HCl rarr 3 ICl + 3 KCl + 3H_(2)O`
According to the equation given, 1 mole of `KIO_(3)` is equivalent to 2 moles of KI
Number of millimoles of KI in 20 mL of stock solution = `2 xx 3 = 6`
Number of millimoles of KI in 50 mL of the same solution = `6 xx 50/20 = 15 `
Number of millimoles of `KIO_(3)` in 50 mL of solution = 1/10 xx 50 = 5
Number of millimoles of KI used with `AgNO_(3)` = 15 - 10 = 5
`AgNO_(3) + KI rarr AgI + KNO_(3)`
1 mole of `AgNO_(3)` reacts with 1 mole of KI. Therefore, number of millimoles of `AgCl_(3)` is equal to 5
Weight of `AgNO_(3) = 5 xx 10^(-3) xx 170 g = 0.85 g`
`% of AgNO_(3) = 0.85 xx 100/1.0 = 85.0 % `