`100 cm^(3)` of a given sample of `H_(2)O_(2)`, gives `1000cm^(3)` of `O_(2)` at S.T.P. The given sample is

Options

A. 10% `H_(2)O_(2)`
B. 90% `H_(2)O_(2)`
C. 10 volume `H_(2)O_(2)`
D. 100 volume `H_(2)O_(2)`

Explanation

To solve the problem, we need to determine the concentration of the hydrogen peroxide (H₂O₂) sample based on the volume of oxygen (O₂) it produces when decomposed.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The decomposition of hydrogen peroxide can be represented by the following balanced chemical 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 H₂O₂ produce 1 mole of O₂.

2. **Volume Relationship at STP**:
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters (or 22400 cm³). Therefore, the volume of oxygen produced can be directly related to the volume of H₂O₂ decomposed.

3. **Given Data**:
- Volume of H₂O₂ = 100 cm³
- Volume of O₂ produced = 1000 cm³

4. **Calculating the Volume Ratio**:
According to the stoichiometry of the reaction:
\[ \text{Volume of O}_2 = \frac{1}{2} \times \text{Volume of H}_2\text{O}_2 \]
Therefore, if 100 cm³ of H₂O₂ produces 1000 cm³ of O₂, we can set up the equation:
\[ \text{Volume of O}_2 = \frac{1000 \text{ cm}^3}{100 \text{ cm}^3} = 10 \]
This means that 1 cm³ of H₂O₂ produces 10 cm³ of O₂.

5. **Determining the Volume Strength**:
The volume strength of H₂O₂ is defined as the volume of O₂ produced by 1 volume of H₂O₂. Since 1 cm³ of H₂O₂ produces 10 cm³ of O₂, the volume strength is 10 volumes.

6. **Identifying the Correct Option**:
Now, we can compare our calculated volume strength with the options given:
- 10% of H₂O₂
- 90% H₂O₂
- 10 volume H₂O₂
- 100 volume H₂O₂

The correct answer is **10 volume H₂O₂**, as this matches our calculated volume strength.

### Final Answer:
The given sample is **10 volume H₂O₂**.