The strength of `H_(2)O_(2)` is expressed in many ways like molarity , normality , % strength and volume strengths But out of all these form of strengths , volume strength has great significance for chemical reactions .
This decomposition of `H_(2)O_(2)` is shown as under
`H_(2)O_(2)(l) to H_(2)O(l) + (1)/(2)O_(2)(g)`
'x' volume strength of `H_(2)O_(2)` means one volume (litre or ml) of `H_(2)O_(2)` releases x volume (litre or ml) of `O_(2)` at NTP .
1 litre `H_(2)O_(2)` release x litre of `O_(2)` at NTP
`=(x)/(22.4)` moles of `O_(2)`
From the equation ,
1 mole of `O_(2)` produces from 2 moles of `H_(2)O_(2)` .
`(x)/(22.4)` moles of `O_(2)` produces from `2xx(x)/(22.4)` moles of `H_(2)O_(2)`
`=(x)/(11.2)` moles of `H_(2)O_(2)`
So, molarity of `H_(2)O_(2)=((x)/(11.2))/(1)=(x)/(11.2)M`
Normality =n-factor `xx` molarity
`=2xx(x)/(11.2)=(x)/(5.6) N`
What is the percentage strength of "15 volume" `H_(2)O_(2)` ?

Options

A. `6.086 %`
B. `4.55%`
C. `3.03%`
D. `1.5%`

Answer

To find the percentage strength of "15 volume" H₂O₂, we will follow these steps:

### Step 1: Understand Volume Strength
Volume strength of H₂O₂ indicates how many volumes of O₂ gas are produced from one volume of H₂O₂ at Normal Temperature and Pressure (NTP). For example, "15 volume" means that 1 liter (or 1 ml) of H₂O₂ releases 15 liters (or 15 ml) of O₂ at NTP.

### Step 2: Relate Volume Strength to Molarity
From the decomposition reaction:
\[ \text{H}_2\text{O}_2 (l) \rightarrow \text{H}_2\text{O} (l) + \frac{1}{2} \text{O}_2 (g) \]
1 mole of H₂O₂ produces 0.5 moles of O₂.

At NTP, 1 mole of gas occupies 22.4 liters. Therefore, 0.5 moles of O₂ will occupy:
\[ 0.5 \times 22.4 \, \text{liters} = 11.2 \, \text{liters} \]

Thus, 1 mole of H₂O₂ produces 11.2 liters of O₂.

### Step 3: Calculate Molarity of H₂O₂
Given that 15 volumes of H₂O₂ produce 15 liters of O₂, we can find the molarity of H₂O₂:
- The amount of O₂ produced from 1 liter of H₂O₂ is 15 liters.
- The number of moles of O₂ produced from 15 liters is:
\[ \frac{15}{22.4} \, \text{moles of O}_2 \]

From the stoichiometry of the reaction, the moles of H₂O₂ required to produce this amount of O₂ is:
\[ 2 \times \frac{15}{22.4} = \frac{30}{22.4} \, \text{moles of H}_2\text{O}_2 \]

Since this is for 1 liter of H₂O₂ solution, the molarity (M) of H₂O₂ is:
\[ M = \frac{30}{22.4} \]

### Step 4: Calculate Molarity Numerically
Calculating the above expression:
\[ M = \frac{30}{22.4} \approx 1.34 \, \text{M} \]

### Step 5: Convert Molarity to Percentage Strength (W/V)
The relationship between molarity and weight/volume percentage (W/V) is given by:
\[ M = \frac{W/V \times 10}{\text{Molar Mass}} \]

The molar mass of H₂O₂ is 34 g/mol. Rearranging the formula to find W/V:
\[ W/V = \frac{M \times \text{Molar Mass}}{10} \]

Substituting the values we have:
\[ W/V = \frac{1.34 \times 34}{10} \]

### Step 6: Calculate W/V Percentage
Calculating the above expression:
\[ W/V = \frac{45.56}{10} = 4.56\% \]

### Conclusion
The percentage strength of "15 volume" H₂O₂ is approximately **4.56%**.

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