The strength of `H_(2)O_(2)` is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of `H_(2)O_(2)` on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of `H_(2)O_(2)`gives 10 litre of `O_(2)` at 1 atm and 273 K The decomposition of `H_(2)O_(2)` is shown as under :
`H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g)`
`H_(2)O_(2)` can acts as oxidising as well as reducing agent. As oxidizing agent `H_(2)O_(2)` is converted into `H_(2)O` and as reducing agent `H_(2)O_(2)` is converted into `O_(2)`. For both cases its n-factor is 2. ` :. "Normality " "of" H_(2)O_(2)` " solution " `=2xx "molarity of" H_(2)O_(2)` solution
40 g `Ba(MnO_(4))_(2)` (mol.mass=375) sample containing some inert impurities in acidic medium completely reacts with 125 mL of "33.6 V" of `H_(2)O_(2)`. What is the percentage purity of the sample ?

Options

A. 0.2812
B. 0.7031
C. 0.85
D. None of these

Answer

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

### Step 1: Understand the Reaction
The reaction between H₂O₂ and Ba(MnO₄)₂ in acidic medium can be represented as follows:
\[ \text{Ba(MnO}_4\text{)}_2 + \text{H}_2\text{O}_2 \rightarrow \text{Ba}^{2+} + 2\text{Mn}^{2+} + \text{O}_2 + \text{H}_2\text{O} \]

### Step 2: Calculate the Molarity of H₂O₂
Given that the strength of H₂O₂ is 33.6 V, we can convert this to molarity. The volume strength indicates that 1 volume of H₂O₂ produces 33.6 volumes of O₂.

Using the molar volume of a gas at STP (22.4 L):
\[ \text{Molarity of H₂O₂} = \frac{33.6 \, \text{L}}{22.4 \, \text{L}} = 1.5 \, \text{mol/L} \]

### Step 3: Calculate the Number of Moles of H₂O₂
Given the volume of H₂O₂ used is 125 mL (or 0.125 L):
\[ \text{Number of moles of H₂O₂} = \text{Molarity} \times \text{Volume} = 1.5 \, \text{mol/L} \times 0.125 \, \text{L} = 0.1875 \, \text{mol} \]

### Step 4: Determine the n-factor of H₂O₂
As H₂O₂ acts as a reducing agent in this reaction, its n-factor is 2 (since it produces O₂).

### Step 5: Calculate the Gram Equivalent of H₂O₂
\[ \text{Gram equivalent of H₂O₂} = \text{Number of moles} \times \text{n-factor} = 0.1875 \, \text{mol} \times 2 = 0.375 \, \text{equivalents} \]

### Step 6: Calculate the Gram Equivalent of Ba(MnO₄)₂
The n-factor for Ba(MnO₄)₂ can be calculated based on the change in oxidation state of Mn from +7 to +2:
\[ \text{n-factor of Ba(MnO₄)₂} = 2 \, \text{(Mn)} \times 5 = 10 \]

### Step 7: Calculate the Number of Moles of Ba(MnO₄)₂
Using the equivalence concept:
\[ \text{Gram equivalent of Ba(MnO₄)₂} = \text{Number of moles} \times \text{n-factor} \]
Setting the gram equivalents equal:
\[ 0.375 = \text{Number of moles of Ba(MnO₄)₂} \times 10 \]
\[ \text{Number of moles of Ba(MnO₄)₂} = \frac{0.375}{10} = 0.0375 \, \text{mol} \]

### Step 8: Calculate the Mass of Ba(MnO₄)₂
Using the molar mass of Ba(MnO₄)₂, which is 375 g/mol:
\[ \text{Mass of Ba(MnO₄)₂} = \text{Number of moles} \times \text{Molar mass} = 0.0375 \, \text{mol} \times 375 \, \text{g/mol} = 14.0625 \, \text{g} \]

### Step 9: Calculate the Percentage Purity
Given that the total mass of the sample is 40 g:
\[ \text{Percentage purity} = \left( \frac{\text{Mass of pure Ba(MnO₄)₂}}{\text{Total mass of sample}} \right) \times 100 = \left( \frac{14.0625 \, \text{g}}{40 \, \text{g}} \right) \times 100 = 35.15625\% \]

### Final Answer
The percentage purity of the sample is approximately **35.16%**.

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