A bottle of an aqueous `H_(2)O_(2)` solution is labelled as '28V' `H_(2)O_(2)` and the density of the solution `("in" g//mL)` is 1.25. Choose the correct option.

Options

A. `"Molarity of" H_(2)O_(2) "solution is" 2 `
B. `"Molarity of" H_(2)O_(2) "solution is" 5 `
C. `"Molarity of" H_(2)O_(2) "solution is" 2.15 `
D. none of these

Answer

To solve the problem, we need to determine the molarity of the hydrogen peroxide (H₂O₂) solution labeled as '28V' with a density of 1.25 g/mL. Here’s a step-by-step solution:

### Step 1: Understand the meaning of '28V' H₂O₂
The label '28V' indicates that 1 cm³ (or 1 mL) of H₂O₂ solution can produce 28 cm³ (or 28 mL) of oxygen gas (O₂) when it decomposes. This is a measure of the volume strength of the solution.

### Step 2: Write the decomposition reaction
The decomposition of hydrogen peroxide can be represented by the following balanced chemical equation:
\[ 2 \text{H}_2\text{O}_2 (aq) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g) \]

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

### Step 3: Calculate the molarity using volume strength
The formula relating volume strength (V) to molarity (M) is:
\[ V = 11.2 \times M \]
Where V is the volume strength in terms of mL of O₂ produced per mL of H₂O₂ solution.

Given that the volume strength is 28:
\[ 28 = 11.2 \times M \]

### Step 4: Solve for molarity (M)
Rearranging the formula to find M:
\[ M = \frac{28}{11.2} \]

Calculating this gives:
\[ M = 2.5 \, \text{mol/L} \]

### Step 5: Conclusion
The molarity of the H₂O₂ solution is 2.5 mol/L. However, if the options provided do not include 2.5, we would select "none of these" as the correct answer.