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
What is thepercentage strength (%w/V) of "11.2 V" `H_(2)O_(2)`

Options

A. 1.7
B. 3.4
C. 34
D. None of these

Answer

To find the percentage strength (% w/V) of "11.2 V" H₂O₂, we can follow these steps:

### Step 1: Understand the meaning of "11.2 V"
The term "11.2 V" means that 1 liter of H₂O₂ will produce 11.2 liters of oxygen gas (O₂) when it decomposes at standard temperature and pressure (STP).

### Step 2: Calculate the number of moles of oxygen produced
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the number of moles of O₂ produced from the volume given:
\[ \text{Number of moles of O₂} = \frac{\text{Volume of O₂}}{\text{Molar volume at STP}} = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{moles} \]

### Step 3: Relate moles of O₂ to moles of H₂O₂
From the decomposition reaction:
\[ \text{H₂O₂} \rightarrow \text{H₂O} + \frac{1}{2} \text{O₂} \]
1 mole of H₂O₂ produces 0.5 moles of O₂. Therefore, if we have 0.5 moles of O₂, the number of moles of H₂O₂ used will be:
\[ \text{Number of moles of H₂O₂} = 2 \times \text{Number of moles of O₂} = 2 \times 0.5 = 1 \, \text{mole} \]

### Step 4: Calculate the mass of H₂O₂
The molecular weight of H₂O₂ is 34 g/mol. Thus, the mass of H₂O₂ can be calculated as:
\[ \text{Mass of H₂O₂} = \text{Number of moles} \times \text{Molecular weight} = 1 \, \text{mole} \times 34 \, \text{g/mol} = 34 \, \text{g} \]

### Step 5: Calculate the percentage strength (% w/V)
The percentage weight by volume (% w/V) is calculated using the formula:
\[ \text{% w/V} = \left( \frac{\text{Weight of solute (g)}}{\text{Volume of solution (mL)}} \right) \times 100 \]
Since we have 34 g of H₂O₂ in 1 liter (or 1000 mL) of solution:
\[ \text{% w/V} = \left( \frac{34 \, \text{g}}{1000 \, \text{mL}} \right) \times 100 = 3.4\% \]

### Final Answer
The percentage strength (% w/V) of "11.2 V" H₂O₂ is **3.4%**.

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