`1 g` sample of `AgNO_(3)` is dissolved in `50 mL` of water, It is titrated with `50 mL` of `KI` solution. The `Agl`percipitated is filtered off. Excess of `KI` filtrate is titrated with `M//10KIO_(3)` in presence of `6 M HCl` till all `I^(-)` converted into `ICI`. It requires `50 mL` of `M//10 KIO_(3)` solution. `20 mL` of the same stock solution of `KI` requires `30 mL` of `M//10KIO_(3)` under similar conditions. Calculate `%` of `AgNO_(3)` in sample. The reaction is
`KIO_(3)+2KI+6HClrarr3ICl+3KCl+3H_(2)O`

Explanation

To solve the problem step by step, we will break down the information given and perform the necessary calculations.

### Step 1: Understand the reactions involved
The reaction between potassium iodate (KIO3), potassium iodide (KI), and hydrochloric acid (HCl) is given as:
\[ \text{KIO}_3 + 2 \text{KI} + 6 \text{HCl} \rightarrow 3 \text{ICl} + 3 \text{KCl} + 3 \text{H}_2\text{O} \]

### Step 2: Analyze the titration data
1. **First titration**: 50 mL of KI solution is used to react with AgNO3. The excess KI is then titrated with KIO3.
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