A `8gm` sample of iron are containing `V%` (by wt) of iron `(Fe^(2+))` was present in `1L` solution. This solution required `V` ml of a `KMnO_(4)` solution for oxidation of `Fe^(2+)` in acidic medium. If in another titration `20ml` of same `KMnO_(4)` is required, for oxidaton of `20ml` of `H_(2)O_(2)` in acidic medium. Find volume strength of `H_(2)O_(2)`

Explanation

To solve the problem, we need to find the volume strength of hydrogen peroxide (H₂O₂) based on the titration data provided. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the milli-equivalents of Fe²⁺
The mass of iron (Fe) in the sample is given as 8 grams. The percentage of Fe²⁺ in the sample is V%.

The formula for calculating milli-equivalents is:
\[ \text{milli-equivalents} = \frac{\text{mass (g)}}{\text{equivalent weight (g/equiv)}} \]

The equivalent weight of Fe (for Fe²⁺) is:
\[ \text{Equivalent weight of Fe} = \frac{56}{2} = 28 \, \text{g/equiv} \]

Thus, the milli-equivalents of Fe²⁺ in the solution can be calculated as:
\[ \text{milli-equivalents of Fe}^{2+} = \frac{8 \times V}{28} \times 1000 = \frac{8000V}{28} = \frac{2000V}{7} \, \text{milli-equivalents} \]

### Step 2: Relate milli-equivalents of Fe²⁺ to KMnO₄
According to the first titration, the milli-equivalents of Fe²⁺ will equal the milli-equivalents of KMnO₄ used, which is V mL of KMnO₄ solution.

Let the normality of KMnO₄ be N. The milli-equivalents of KMnO₄ can also be expressed as:
\[ \text{milli-equivalents of KMnO}_4 = N \times V \]

Setting these equal gives:
\[ \frac{2000V}{7} = N \times V \]

### Step 3: Solve for the normality of KMnO₄
Dividing both sides by V (assuming V ≠ 0):
\[ N = \frac{2000}{7} \, \text{N} \]

### Step 4: Analyze the second titration with H₂O₂
In the second titration, 20 mL of KMnO₄ is required to oxidize 20 mL of H₂O₂. The milli-equivalents of KMnO₄ used here can be expressed as:
\[ \text{milli-equivalents of KMnO}_4 = N \times 20 \]

Substituting the value of N:
\[ \text{milli-equivalents of KMnO}_4 = \frac{2000}{7} \times 20 = \frac{40000}{7} \, \text{milli-equivalents} \]

### Step 5: Relate milli-equivalents of H₂O₂
The milli-equivalents of H₂O₂ can be expressed as:
\[ \text{milli-equivalents of H}_2O_2 = \frac{20 \times \text{Volume Strength}}{5.6} \]

Setting the milli-equivalents of KMnO₄ equal to the milli-equivalents of H₂O₂ gives:
\[ \frac{40000}{7} = \frac{20 \times \text{Volume Strength}}{5.6} \]

### Step 6: Solve for Volume Strength of H₂O₂
Cross-multiplying to solve for Volume Strength:
\[ 40000 \times 5.6 = 20 \times \text{Volume Strength} \times 7 \]
\[ 224000 = 140 \times \text{Volume Strength} \]
\[ \text{Volume Strength} = \frac{224000}{140} = 1600 \]

### Step 7: Final Calculation
To find the volume strength, we divide by 100:
\[ \text{Volume Strength} = 8 \]

Thus, the volume strength of H₂O₂ is **8**.

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