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
20mL of `H_(2)O_(2)` solution is reacted with 80 mL of 0.05 `MKMnO_(4)` "in acidic medium then what is the volume strength of" `H_(2)O_(2)` ?

Options

A. 2.8
B. 5.6
C. 11.2
D. None of these

Answer

To find the volume strength of \( H_2O_2 \) from the given data, we can follow these steps:

### Step 1: Understand the Reaction
The decomposition of \( H_2O_2 \) can be represented as:
\[ H_2O_2(aq) \rightarrow H_2O(l) + \frac{1}{2}O_2(g) \]
In this reaction, one mole of \( H_2O_2 \) produces half a mole of \( O_2 \).

### Step 2: Determine the n-factor
The n-factor for \( H_2O_2 \) is 2 when it acts as both an oxidizing and reducing agent. This means that 1 mole of \( H_2O_2 \) can either provide 1 mole of \( O_2 \) or consume 2 moles of electrons.

### Step 3: Calculate the number of equivalents of \( KMnO_4 \)
Given that we have 80 mL of \( 0.05 \, M \, KMnO_4 \):
- Molarity (M) = 0.05 mol/L
- Volume (V) = 80 mL = 0.08 L

The number of equivalents of \( KMnO_4 \) can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume} \times n \]
For \( KMnO_4 \), the n-factor is 5 (since it gains 5 electrons):
\[ \text{Number of equivalents of } KMnO_4 = 0.05 \, \text{mol/L} \times 0.08 \, \text{L} \times 5 = 0.02 \, \text{equivalents} \]

### Step 4: Relate the equivalents of \( H_2O_2 \) to \( KMnO_4 \)
Since the number of equivalents of \( H_2O_2 \) must equal the number of equivalents of \( KMnO_4 \):
\[ \text{Number of equivalents of } H_2O_2 = 0.02 \, \text{equivalents} \]

### Step 5: Calculate the Normality of \( H_2O_2 \)
Let the normality of \( H_2O_2 \) be \( N \) and the volume of \( H_2O_2 \) used is 20 mL = 0.02 L.
Using the formula for equivalents:
\[ \text{Number of equivalents} = N \times \text{Volume (L)} \]
So,
\[ 0.02 = N \times 0.02 \]
Thus,
\[ N = 1 \, N \]

### Step 6: Calculate the Volume Strength of \( H_2O_2 \)
The volume strength of \( H_2O_2 \) is calculated using the formula:
\[ \text{Volume Strength} = 5.6 \times \text{Normality} \]
Substituting the normality we found:
\[ \text{Volume Strength} = 5.6 \times 1 = 5.6 \]

### Final Answer
The volume strength of \( H_2O_2 \) is **5.6 V**.

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