A substance ‘A’ decomposes in solution following first-order kinetics. Flask 1 contains 1 l of 1 M solution of ‘A’ and flask 2 contains 100 ml of 0.6 M solution of ‘A’. After 8.0 h, the concentration of ‘A’ in flask 1 becomes 0.25 M. In what time, the concentration of ‘A’ in flask 2 becomes 0.3 M?

4.0 h

The correct answer is option 3. 4.0 h.

To solve this problem, we can use the first-order kinetics equation:

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

Where \([A]_t\) is the concentration of \(A\) at time \(t\), \([A]_0\) is the initial concentration of \(A\), \(k\) is the rate constant, and \(t\) is the time.

Let's first calculate the rate constant (\(k\)) using the information from flask 1:

\([A]_t = 0.25 \, \text{M}\)

\([A]_0 = 1 \, \text{M}\)

\(t = 8.0 \, \text{h}\)

\(\ln\left(\frac{{0.25}}{{1}}\right) = -k \times 8.0\)

\(-1.3863 = -8k\)

\(k \approx 0.1733 \, \text{h}^{-1}\)

Now, let's calculate the time (\(t\)) required for the concentration of \(A\) in flask 2 to become 0.3 M:

\([A]_t = 0.3 \, \text{M}\)

\([A]_0 = 0.6 \, \text{M}\)

\(k = 0.1733 \, \text{h}^{-1}\)

\(\ln\left(\frac{{0.3}}{{0.6}}\right) = -0.1733 \times t\)

\(-0.6931 = -0.1733 \times t\)

\(t \approx 4.0 \, \text{h}\)

Therefore, the concentration of \(A\) in flask 2 becomes 0.3 M after approximately 4.0 hours. The correct answer is (3) 4.0 h.