The volume strength of `"8.9 M "H_(2)O_(2)` solution calculated at 273 K and 1 atm is ______. `("R = 0.0821 L atm K"^(-1)" mol"^(-1))` (rounded off ot the nearest integer)

Explanation

To calculate the volume strength of an 8.9 M H₂O₂ solution at standard conditions (273 K and 1 atm), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between molarity and normality**:
- The normality (N) of a solution is related to its molarity (M) by the equation:
\[ N = \frac{M}{n} \]
- For hydrogen peroxide (H₂O₂), the n-factor is 2 because it can donate two moles of electrons (or two moles of H⁺ ions).

2. **Calculate the normality of the H₂O₂ solution**:
- Given that the molarity (M) of the solution is 8.9 M:
\[ N = \frac{8.9}{2} = 4.45 \, \text{N} \]

3. **Relate normality to volume strength**:
- 1 normal H₂O₂ solution corresponds to 5.6 volumes of oxygen gas at standard conditions.
- Therefore, the volume strength (V) can be calculated as:
\[ V = N \times 5.6 \]

4. **Calculate the volume strength**:
- Substitute the normality into the equation:
\[ V = 4.45 \times 5.6 = 24.92 \, \text{volumes} \]

5. **Calculate the volume strength for the given molarity**:
- Since we need to find the volume strength for an 8.9 M solution, we can multiply the volume strength of 1 N solution by the factor of 8.9:
\[ V_{8.9} = 5.6 \times 8.9 = 49.68 \, \text{volumes} \]

6. **Round off to the nearest integer**:
- Rounding 49.68 gives us approximately 50.

### Final Answer:
The volume strength of the 8.9 M H₂O₂ solution is **50**.