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 273 K and 1 atm, we can follow these steps:

### Step 1: Understand the decomposition of H₂O₂
When hydrogen peroxide (H₂O₂) decomposes, it produces water (H₂O) and oxygen gas (O₂). The balanced equation for this decomposition is:
\[ 2 \, \text{H}_2\text{O}_2 \rightarrow 2 \, \text{H}_2\text{O} + \text{O}_2 \]

From the equation, we can see that:
- 2 moles of H₂O₂ produce 1 mole of O₂.

### Step 2: Relate volume strength to molarity
The volume strength of a solution is defined as the volume of oxygen gas produced (in liters) when 1 mole of H₂O₂ decomposes. From the ideal gas law, we know that at standard temperature and pressure (STP: 273 K and 1 atm), 1 mole of any gas occupies 22.4 liters.

Since 2 moles of H₂O₂ produce 1 mole of O₂, we can conclude:
- 1 mole of H₂O₂ produces 0.5 moles of O₂.
- Therefore, 1 mole of H₂O₂ produces \( 0.5 \times 22.4 \, \text{L} = 11.2 \, \text{L} \) of O₂.

### Step 3: Calculate the volume strength
The volume strength (VS) can be calculated using the formula:
\[ \text{Volume Strength} = \text{Molarity} \times 11.2 \]

Substituting the given molarity of H₂O₂:
\[ \text{Volume Strength} = 8.9 \, \text{M} \times 11.2 \, \text{L} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ \text{Volume Strength} = 8.9 \times 11.2 = 99.68 \]

### Step 5: Round off to the nearest integer
Rounding off 99.68 gives us:
\[ \text{Volume Strength} \approx 100 \]

### Final Answer
The volume strength of the 8.9 M H₂O₂ solution is **100**.

---