A commercial sample of hydrogen peroxide is labelled as 10 volume. Its percentage strength is nearly

Options

A. `3%`
B. `1%`
C. `90%`
D. `10%`

Answer

To solve the problem of determining the percentage strength of a commercial sample of hydrogen peroxide labeled as 10 volume, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Volume Label**:
- The term "10 volume" means that 1 volume of hydrogen peroxide can produce 10 volumes of oxygen gas (O2) at Normal Temperature and Pressure (NTP).

2. **Reaction of Hydrogen Peroxide**:
- The decomposition of hydrogen peroxide can be represented by the equation:
\[ 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \]
- From this equation, we see that 2 moles of hydrogen peroxide produce 1 mole of oxygen.

3. **Volume of Oxygen Produced**:
- At NTP, 1 mole of gas occupies 22,400 mL. Therefore, 1 mole of O2 will occupy 22,400 mL.
- Since 10 volumes of O2 are produced from 1 volume of H2O2, we can say:
\[ 10 \text{ volumes of O}_2 = 10 \times 22,400 \text{ mL} = 224,000 \text{ mL} \]

4. **Calculating Moles of H2O2**:
- To produce 224,000 mL of O2, we need:
\[ \text{Moles of O}_2 = \frac{224,000 \text{ mL}}{22,400 \text{ mL/mole}} = 10 \text{ moles of O}_2 \]
- Since 2 moles of H2O2 produce 1 mole of O2, to produce 10 moles of O2, we need:
\[ \text{Moles of H}_2\text{O}_2 = 10 \times 2 = 20 \text{ moles of H}_2\text{O}_2 \]

5. **Calculating Mass of H2O2**:
- The molar mass of H2O2 (hydrogen peroxide) is approximately 34 g/mol.
- Therefore, the mass of 20 moles of H2O2 is:
\[ \text{Mass of H}_2\text{O}_2 = 20 \text{ moles} \times 34 \text{ g/mol} = 680 \text{ grams} \]

6. **Calculating Percentage Strength**:
- The percentage strength of a solution is calculated as:
\[ \text{Percentage Strength} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100 \]
- Assuming the solution is 1 liter (1000 mL) of water (which is approximately 1000 grams), the total mass of the solution is:
\[ \text{Mass of solution} = 1000 \text{ g (water)} + 680 \text{ g (H}_2\text{O}_2) = 1680 \text{ g} \]
- Now, substituting into the percentage formula:
\[ \text{Percentage Strength} = \left( \frac{680}{1680} \right) \times 100 \approx 40.48\% \]

### Final Answer:
The percentage strength of the commercial sample of hydrogen peroxide labeled as 10 volume is approximately **40.48%**.