0.31 gm of an alloy Fe+ Cu was dissolved in excess dilute `H_(2)SO_(4)` and the solution was made up to 100 ml . 20 ml of this solution required 3 ml of `N/30 K_(2)Cr_(2)O_(7)` solution for exact oxidation. The % purity (in closest value) of Fe in wire is ____________.

Answer

To solve the problem step by step, we will follow the outlined process to determine the percentage purity of iron in the alloy.

### Step 1: Understanding the Reaction
When the alloy of iron (Fe) and copper (Cu) is dissolved in dilute sulfuric acid (H₂SO₄), only iron reacts to form ferrous sulfate (FeSO₄), while copper remains unreacted.

### Step 2: Titration Data
We have the following data from the problem:
- Total mass of the alloy = 0.31 g
- Volume of the solution made = 100 ml
- Volume of the titrated solution = 20 ml
- Volume of K₂Cr₂O₇ used = 3 ml
- Normality of K₂Cr₂O₇ = N/30

### Step 3: Calculate the Milliequivalents of K₂Cr₂O₇
To find the milliequivalents of K₂Cr₂O₇ used in the titration, we use the formula:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (ml)} \]
Substituting the values:
\[ \text{Milliequivalents of K₂Cr₂O₇} = \frac{1}{30} \times 3 = \frac{3}{30} = \frac{1}{10} \text{ equivalents} \]

### Step 4: Calculate the Moles of K₂Cr₂O₇
Since 1 mole of K₂Cr₂O₇ reacts with 6 moles of FeSO₄, we need to convert the milliequivalents to moles:
\[ \text{Moles of K₂Cr₂O₇} = \frac{1/10}{6} = \frac{1}{60} \text{ moles} \]

### Step 5: Calculate the Moles of FeSO₄ in 20 ml
Using the stoichiometry of the reaction:
\[ \text{Moles of FeSO₄} = \text{Moles of K₂Cr₂O₇} \times 6 = \frac{1}{60} \times 6 = \frac{1}{10} \text{ moles} \]

### Step 6: Calculate the Moles of FeSO₄ in 100 ml
Since the total solution is 100 ml, we can find the total moles of FeSO₄ in the entire solution:
\[ \text{Moles of FeSO₄ in 100 ml} = \frac{1}{10} \times 5 = \frac{5}{10} = 0.5 \text{ moles} \]

### Step 7: Convert Moles of FeSO₄ to Mass of Iron
The molar mass of iron (Fe) is approximately 56 g/mol. Therefore, the mass of iron in the solution is:
\[ \text{Mass of Fe} = \text{Moles of FeSO₄} \times \text{Molar mass of Fe} = 0.5 \times 56 = 28 \text{ grams} \]

### Step 8: Calculate the Percentage Purity of Iron in the Alloy
To find the percentage purity of iron in the alloy, we use the formula:
\[ \text{Percentage purity of Fe} = \left(\frac{\text{Mass of Fe}}{\text{Mass of alloy}}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage purity of Fe} = \left(\frac{28}{0.31}\right) \times 100 \approx 9\% \]

### Final Answer
The percentage purity of iron in the alloy is **9%**.

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