A substance `''A''` decomposes in solution following the first order kinetics. Flask `I` contains `L` of `1 M` solution of `A` and falsk `II` constains `100 mL` of `0.6 M` solution. After `8 hr`, the concentration, of `A` in flask `I` becomes `0.25 M`. What will be the time for concentration of `A` in flask `II` to become `0.3 M` ?

Options

A. `0.4 hr`
B. `2.4 hr`
C. `4.0 hr`
D. Unpredictable as rate constant is not given

Explanation

To solve the problem, we will use the first-order kinetics equation and the information provided about the two flasks.

### Step-by-Step Solution:

**Step 1: Understand the first-order kinetics equation.**
For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The integrated rate law for a first-order reaction is given by:

\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]

Where:
- \([A]_0\) = initial concentration of A
- \([A]\) = concentration of A at time \(t\)
- \(k\) = rate constant
- \(t\) = time

**Step 2: Calculate the rate constant \(k\) for Flask I.**
In Flask I, the initial concentration \([A]_0\) is 1 M and the concentration after 8 hours \([A]\) is 0.25 M.

Using the first-order equation:

\[ \ln \left( \frac{1}{0.25} \right) = k \cdot 8 \]

Calculating the left side:

\[ \ln(4) = k \cdot 8 \]

Now, we know that \(\ln(4) \approx 1.386\):

\[ 1.386 = k \cdot 8 \]

Solving for \(k\):

\[ k = \frac{1.386}{8} \approx 0.17325 \, \text{hr}^{-1} \]

**Step 3: Use the rate constant to find the time for Flask II.**
In Flask II, the initial concentration \([A]_0\) is 0.6 M and we want to find the time \(t\) when the concentration becomes 0.3 M.

Using the first-order equation again:

\[ \ln \left( \frac{0.6}{0.3} \right) = k \cdot t \]

Calculating the left side:

\[ \ln(2) = k \cdot t \]

Knowing that \(\ln(2) \approx 0.693\):

\[ 0.693 = 0.17325 \cdot t \]

Solving for \(t\):

\[ t = \frac{0.693}{0.17325} \approx 4.00 \, \text{hours} \]

### Final Answer:
The time for the concentration of A in Flask II to become 0.3 M is approximately **4 hours**.