For the decomposition of `H_2O_2(aq)` it was found that `V_(O_2)` (t=15 min) was 100 mL (at `0^@C` and 1 atm) while `V_(O_2)` (maximum) was 200 mL (at `0^@C` and 2 atm). If the same reaction had been followed by the titration method of if `V_(KMnO_(4))^((cM))`(t=0) had been 40 mL, what would `V_(KMnO_(4))^((cM))`(t=15 min) have been ?

Options

A. 30 mL
B. 25 mL
C. 20 mL
D. 15 mL

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
- Volume of O₂ at t = 15 min (V₁) = 100 mL (at 1 atm)
- Maximum volume of O₂ (V_max) = 200 mL (at 2 atm)
- Initial volume of KMnO₄ (V(KMnO₄)₀) = 40 mL

### Step 2: Use Boyle's Law to Find the Maximum Volume of O₂ at 1 atm
According to Boyle's law, at constant temperature:
\[ P_1 V_1 = P_2 V_2 \]

Here:
- P₁ = 1 atm (initial pressure)
- V₁ = Volume of O₂ at 1 atm (to be calculated)
- P₂ = 2 atm (maximum pressure)
- V₂ = 200 mL (maximum volume of O₂)

Using the formula:
\[ 1 \, \text{atm} \cdot V_1 = 2 \, \text{atm} \cdot 200 \, \text{mL} \]

Rearranging gives:
\[ V_1 = \frac{2 \, \text{atm} \cdot 200 \, \text{mL}}{1 \, \text{atm}} = 400 \, \text{mL} \]

### Step 3: Determine the Fraction of Reaction Completed
We know that at t = 15 min, the volume of O₂ produced is 100 mL. The maximum volume we calculated is 400 mL.

The fraction of the reaction that is completed can be calculated as:
\[ \text{Fraction completed} = \frac{V_{O₂ \, \text{produced}}}{V_{O₂ \, \text{max}}} = \frac{100 \, \text{mL}}{400 \, \text{mL}} = \frac{1}{4} \]

### Step 4: Determine the Remaining Fraction of the Reaction
If \(\frac{1}{4}\) of the reaction is completed, then the remaining fraction is:
\[ \text{Fraction remaining} = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 5: Calculate the Volume of KMnO₄ at t = 15 min
Since the initial volume of KMnO₄ is 40 mL, the volume at t = 15 min can be calculated as:
\[ V(KMnO₄)_{15 \, \text{min}} = \text{Fraction remaining} \times V(KMnO₄)_{0} \]
\[ V(KMnO₄)_{15 \, \text{min}} = \frac{3}{4} \times 40 \, \text{mL} = 30 \, \text{mL} \]

### Final Answer
The volume of KMnO₄ at t = 15 min is **30 mL**.

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