34 g of `H_(2)O_(2)` is present in 1120 " mL of " solution. This solution is called

Options

A. 10 vol solution
B. 20 vol solution
C. 34 vol solution
D. 32 vol solution

Explanation

To determine the type of solution represented by 34 g of H₂O₂ in 1120 mL of solution, we can follow these steps:

### Step 1: Calculate the Equivalent Weight of H₂O₂
1. **Find the Molar Mass of H₂O₂**:
- The molar mass of H₂O₂ (Hydrogen Peroxide) is calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- O: 16 g/mol × 2 = 32 g/mol
- Total Molar Mass = 2 + 32 = 34 g/mol

2. **Determine the Equivalent Weight**:
- The equivalent weight of a substance is its molar mass divided by the n-factor.
- For H₂O₂, the n-factor is 2 (since it can donate 2 moles of H⁺ ions).
- Equivalent Weight = Molar Mass / n-factor = 34 g/mol / 2 = 17 g/equiv.

### Step 2: Calculate Normality of the Solution
1. **Use the Normality Formula**:
- Normality (N) = (Given weight in grams × 1000) / (Equivalent weight × Volume in mL)
- Substitute the values:
- Given weight = 34 g
- Equivalent weight = 17 g/equiv.
- Volume = 1120 mL

2. **Calculate Normality**:
\[ N = \frac{34 \, \text{g} \times 1000}{17 \, \text{g/equiv.} \times 1120 \, \text{mL}} = \frac{34000}{19040} \approx 1.79 \, \text{N} \]

### Step 3: Convert Normality to Volume Strength
1. **Use the Conversion Factor**:
- 1 Normality of H₂O₂ = 5.6 Volume Strength
- Therefore, to find the volume strength corresponding to the calculated normality:
\[ \text{Volume Strength} = \text{Normality} \times 5.6 = 1.79 \, \text{N} \times 5.6 \approx 10 \, \text{Volume} \]

### Conclusion
The solution containing 34 g of H₂O₂ in 1120 mL is a **10 volume solution**.

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