`H_2 O_2` is reported to be 3.03% by mass. The strength of `H_2 O_2` in terms of volume of `O_2` if density of `H_2 O_2` solution is 'd' g `ml^(-1)` will be:

Options

A. 5d' volume
B. 10d' volume
C. 30d' volume
D. 20d' volume

Answer

To solve the problem, we need to find the strength of hydrogen peroxide (H₂O₂) in terms of the volume of oxygen (O₂) it can produce when heated. The given information includes the mass percentage of H₂O₂ in the solution and its density.

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- Mass percentage of H₂O₂ = 3.03%
- Density of H₂O₂ solution = 'd' g/ml

2. **Assuming the Mass of the Solution:**
- Let's assume we have 100 g of the solution for simplicity.

3. **Calculating the Mass of H₂O₂:**
- Mass of H₂O₂ in the solution = \( \frac{3.03}{100} \times 100 \) g = 3.03 g

4. **Finding the Volume of the Solution:**
- Volume of the solution = \( \frac{\text{mass}}{\text{density}} = \frac{100 \text{ g}}{d \text{ g/ml}} = \frac{100}{d} \) ml

5. **Calculating Moles of H₂O₂:**
- Molecular weight of H₂O₂ = \( 2 \times 1 + 2 \times 16 = 34 \) g/mol
- Moles of H₂O₂ = \( \frac{\text{mass}}{\text{molecular weight}} = \frac{3.03 \text{ g}}{34 \text{ g/mol}} \)

6. **Calculating Molarity of H₂O₂:**
- Molarity (M) = \( \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)
- Volume of solution in liters = \( \frac{100}{d} \times \frac{1}{1000} = \frac{100}{1000d} = \frac{1}{10d} \) L
- Molarity = \( \frac{\frac{3.03}{34}}{\frac{1}{10d}} = \frac{3.03 \times 10d}{34} \)

7. **Calculating Volume Strength:**
- Volume strength = \( 11.2 \times \text{Molarity} \)
- Volume strength = \( 11.2 \times \frac{3.03 \times 10d}{34} \)

8. **Simplifying the Expression:**
- Volume strength = \( \frac{11.2 \times 3.03 \times 10d}{34} \)
- Calculate \( 11.2 \times 3.03 \approx 34.056 \)
- Thus, Volume strength = \( \frac{34.056 \times 10d}{34} = 10d \)

### Final Answer:
The strength of H₂O₂ in terms of volume of O₂ is **10d**.