`2H_(2) O_(2) (l) rarr 2H_(2)o(l) + O_(2) (g)`
`100 mL` of `X` molar `H_(2)O_(2)` gives `3L` of `O_(2)` gas under the condition when 1 moe occupies `24 L`. The value of `X` is

Options

A. 2.5
B. 1
C. 0.5
D. 0.25

Explanation

To find the value of \( X \) in the reaction

\[ 2H_2O_2 (l) \rightarrow 2H_2O (l) + O_2 (g) \]

given that 100 mL of \( X \) molar \( H_2O_2 \) produces 3 L of \( O_2 \) gas under conditions where 1 mole occupies 24 L, we can follow these steps:

### Step 1: Determine the moles of \( O_2 \)
Since we know that 1 mole of gas occupies 24 L, we can calculate the moles of \( O_2 \) produced from 3 L.

\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume}} = \frac{3 \text{ L}}{24 \text{ L/mol}} = \frac{1}{8} \text{ moles} \]

### Step 2: Relate moles of \( O_2 \) to moles of \( H_2O_2 \)
From the balanced equation, we see that 2 moles of \( H_2O_2 \) produce 1 mole of \( O_2 \). Therefore, the moles of \( H_2O_2 \) used can be calculated as follows:

\[ \text{Moles of } H_2O_2 = 2 \times \text{Moles of } O_2 = 2 \times \frac{1}{8} = \frac{1}{4} \text{ moles} \]

### Step 3: Calculate the molarity of \( H_2O_2 \)
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters. We have 100 mL of \( H_2O_2 \), which is equivalent to 0.1 L.

\[ \text{Molarity of } H_2O_2 = \frac{\text{Moles of } H_2O_2}{\text{Volume in L}} = \frac{\frac{1}{4} \text{ moles}}{0.1 \text{ L}} = \frac{1}{4} \div 0.1 = \frac{1}{4} \times 10 = 2.5 \text{ M} \]

### Conclusion
Thus, the value of \( X \) is

\[ \boxed{2.5} \]

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