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

One gram of commercial AgNO_(3) is dissolved in 50 mL of water. It is

1.	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
Answer» Solution :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 % `