A 20.0 mL solution containing 0.2 impure `H_2O_2` reacts completely with 0.316 g of `KMnO_4` in acid solution . The purity of `H_2O_2` (in%) is ...........(mol .wt.of `H_2O_2=34` , mol . Wt . Of `KMnO_4 = 158` )

Answer

To solve the problem, we need to determine the purity of the impure hydrogen peroxide (H₂O₂) solution based on its reaction with potassium permanganate (KMnO₄). Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the moles of KMnO₄ used
We start by calculating the number of moles of KMnO₄ that reacted. The formula to calculate moles is:

\[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Given:
- Mass of KMnO₄ = 0.316 g
- Molar mass of KMnO₄ = 158 g/mol

Calculating moles of KMnO₄:

\[ \text{Moles of KMnO₄} = \frac{0.316 \, \text{g}}{158 \, \text{g/mol}} = 0.00200 \, \text{mol} \]

### Step 2: Determine the equivalent moles of KMnO₄
In acidic medium, KMnO₄ acts as a strong oxidizing agent and undergoes a reduction from Mn(VII) to Mn(II). The change in oxidation state involves the transfer of 5 electrons. Therefore, the equivalent factor for KMnO₄ is 5.

\[ \text{Equivalent moles of KMnO₄} = \text{Moles of KMnO₄} \times \text{Equivalent factor} = 0.00200 \times 5 = 0.01000 \, \text{equivalents} \]

### Step 3: Relate the equivalents of KMnO₄ to H₂O₂
The reaction between KMnO₄ and H₂O₂ can be represented as:

\[ \text{H₂O₂} + \text{KMnO₄} \rightarrow \text{Mn}^{2+} + \text{O}_2 + \text{H₂O} \]

In this reaction, 1 mole of H₂O₂ reacts with 5 equivalents of KMnO₄. Therefore, we can find the moles of H₂O₂ that reacted:

\[ \text{Moles of H₂O₂} = \frac{\text{Equivalent moles of KMnO₄}}{2} = \frac{0.01000}{2} = 0.00500 \, \text{mol} \]

### Step 4: Calculate the weight of pure H₂O₂
Now, we can calculate the weight of pure H₂O₂ that corresponds to the moles we just calculated.

Using the molar mass of H₂O₂ (34 g/mol):

\[ \text{Weight of pure H₂O₂} = \text{Moles of H₂O₂} \times \text{Molar mass of H₂O₂} = 0.00500 \, \text{mol} \times 34 \, \text{g/mol} = 0.17 \, \text{g} \]

### Step 5: Calculate the purity of H₂O₂
The purity of H₂O₂ can be calculated using the formula:

\[ \text{Purity (\%)} = \left( \frac{\text{Weight of pure H₂O₂}}{\text{Weight of impure H₂O₂}} \right) \times 100 \]

Given that the weight of the impure H₂O₂ solution is 0.2 g:

\[ \text{Purity (\%)} = \left( \frac{0.17 \, \text{g}}{0.2 \, \text{g}} \right) \times 100 = 85\% \]

### Final Answer
The purity of H₂O₂ in the solution is **85%**.

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