0.2828 g of iron wire was dissolved in excess dilute `H_(2)SO_(4)` and the solution was made upto 100mL. 20mL of this solution required 30mL of `N/(30)K_(2)Cr_(2)O_(7)` solution for exact oxidation. Calculate percent purity of Fe in wire.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reactions Involved
The iron wire is oxidized in the presence of dilute sulfuric acid (`H2SO4`) to form `Fe^2+`. The `Fe^2+` ions are then oxidized to `Fe^3+` by the dichromate ions from `K2Cr2O7`.

### Step 2: Write the Relevant Reactions
1. **Oxidation of Iron:**
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \]
2. **Oxidation of Fe²⁺ by Dichromate:**
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

### Step 3: Calculate the Milliequivalents of K2Cr2O7
Given:
- Volume of `K2Cr2O7` solution used = 30 mL
- Normality of `K2Cr2O7` = \( \frac{N}{30} \)

The milliequivalents (mEq) of `K2Cr2O7` can be calculated as:
\[ \text{mEq of } K2Cr2O7 = \text{Volume (mL)} \times \text{Normality} = 30 \times \frac{1}{30} = 1 \text{ mEq} \]

### Step 4: Relate mEq of Fe²⁺ to mEq of K2Cr2O7
Since the mEq of `Fe²⁺` in 20 mL of the solution is equal to the mEq of `K2Cr2O7`, we have:
\[ \text{mEq of } Fe^{2+} \text{ in 20 mL} = 1 \text{ mEq} \]

### Step 5: Calculate mEq of Fe²⁺ in 100 mL
To find the mEq of `Fe²⁺` in the entire 100 mL solution:
\[ \text{mEq of } Fe^{2+} \text{ in 100 mL} = \frac{1 \text{ mEq}}{20 \text{ mL}} \times 100 \text{ mL} = 5 \text{ mEq} \]

### Step 6: Calculate Millimoles of Fe²⁺
Using the valency factor (which is 1 for `Fe²⁺`), the millimoles of `Fe²⁺` can be calculated as:
\[ \text{Millimoles of } Fe^{2+} = \text{mEq} \div \text{valency factor} = 5 \div 1 = 5 \text{ mmol} \]

### Step 7: Calculate the Mass of Pure Iron
To find the mass of pure iron:
\[ \text{Mass of pure Fe} = \text{Millimoles} \times \text{Molar mass of Fe} = 5 \times 10^{-3} \text{ mol} \times 56 \text{ g/mol} = 0.28 \text{ g} \]

### Step 8: Calculate Percent Purity of Iron in the Wire
Finally, the percent purity of iron in the wire is calculated as:
\[ \text{Percent purity} = \left( \frac{\text{Mass of pure Fe}}{\text{Mass of wire}} \right) \times 100 = \left( \frac{0.28 \text{ g}}{0.2828 \text{ g}} \right) \times 100 \approx 99\% \]

### Final Answer
The percent purity of Fe in the wire is approximately **99%**.