1. Interpret the graph

For the decomposition reaction $\mathrm{A} \to \mathrm{B}$, the given graph is the standard straight-line plot for a first-order reaction, i.e.

$\log [A] = \log [A]_0 - \frac{k}{2.303}t$

So, from the graph, the slope gives:

$\text{slope} = -\frac{k}{2.303}$

From the graph, the magnitude of slope is read as approximately:

$\frac{k}{2.303} = 0.0301\ \text{s}^{-1}$

Hence,

$k = 2.303 \times 0.0301 \approx 0.0693\ \text{s}^{-1}$

2. Use first-order integrated rate law

Given:

• Initial concentration: $[A]_0 = 50\ \text{g L}^{-1}$
• Final concentration: $[A] = 2.5\ \text{g L}^{-1}$

For a first-order reaction,

$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$

Substitute values:

$t = \frac{2.303}{0.0693} \log \frac{50}{2.5}$

Now,

$\frac{50}{2.5} = 20$

and

$\log 20 = \log(2 \times 10) = \log 2 + 1 = 0.3010 + 1 = 1.3010$

Therefore,

$t = \frac{2.303}{0.0693} \times 1.3010$

Since

$\frac{2.303}{0.0693} \approx 33.23$

we get

$t \approx 33.23 \times 1.3010 \approx 43.2\ \text{s}$

3. Nearest integer

$t \approx 43\ \text{s}$

4. Comparison with stored correct answer

Derived answer = $43$

Stored correct answer = $43$

So the answer matches.