Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation
`2H_(2)O_(2)(aq)rarr2H_(2)O(l)+O_(2)(g)`
under conditions where 1 mole of gas occupies 24 `dm^(3)`. `100cm^(3)` of XM solution of `H_(2)O_(2)` produces 3 `dm^(3)` of `O_(2)`. Thus, X is :

Options

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

Answer

To solve the problem, we need to determine the molarity (X) of the hydrogen peroxide (H₂O₂) solution based on the given reaction and the volume of oxygen produced. Here’s a step-by-step solution:

### Step 1: Write the balanced chemical equation
The decomposition of hydrogen peroxide is given by the equation:
\[ 2H_2O_2 (aq) \rightarrow 2H_2O (l) + O_2 (g) \]

### Step 2: Understand the gas volume relation
We are told that 1 mole of gas occupies 24 dm³. Therefore, we can find out how many moles of oxygen gas (O₂) are produced when 3 dm³ of O₂ is generated.

### Step 3: Calculate moles of O₂ produced
Using the volume of O₂ produced:
\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Volume occupied by 1 mole}} = \frac{3 \, \text{dm}^3}{24 \, \text{dm}^3} = \frac{1}{8} \, \text{moles} \]

### Step 4: Relate moles of O₂ to moles of H₂O₂
From the balanced equation, we see that 2 moles of H₂O₂ produce 1 mole of O₂. Therefore, if we have \(\frac{1}{8}\) moles of O₂, we can find the moles of H₂O₂:
\[ \text{Moles of } H_2O_2 = 2 \times \text{Moles of } O_2 = 2 \times \frac{1}{8} = \frac{1}{4} \, \text{moles} \]

### Step 5: Use the molarity formula
We know that molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]

Here, we have:
- Moles of H₂O₂ = \(\frac{1}{4}\)
- Volume of solution = 100 cm³ = 0.1 L

Substituting these values into the molarity formula:
\[ X = \frac{\frac{1}{4}}{0.1} = \frac{1}{4} \times 10 = 2.5 \, \text{mol/dm}^3 \]

### Conclusion
Thus, the molarity (X) of the hydrogen peroxide solution is:
\[ \boxed{2.5} \]

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