To determine the time required for the concentration of A to decrease from an initial value of $50 \, \mathrm{g} \, \mathrm{L}^{-1}$ to $2.5 \, \mathrm{g} \, \mathrm{L}^{-1}$ in the reaction $ \mathrm{A} \rightarrow \mathrm{B} $, we assume first-order kinetics. Although the graph does not provide a clear indication of the reaction order over the intervals $0-5$, $5-10$, and $10-15$ seconds, where the order appears to be zero, we'll proceed with the assumption of first-order kinetics, since the graph is not a straight line.

The rate constant $ \mathrm{K} $ for first-order reactions can be calculated using the formula:

$ \mathrm{K} = \frac{1}{\mathrm{t}} \ln \frac{\mathrm{A}_0}{\mathrm{A}_{\mathrm{t}}} $

For the interval where $\mathrm{A}_0 = 40 \, \mathrm{g/L}$ and $\mathrm{A}_{\mathrm{t}} = 20 \, \mathrm{g/L}$ after 10 seconds:

$ \mathrm{K} = \frac{1}{10} \ln \frac{40}{20} $

Now, to find the time $ \mathrm{t} $ required for the concentration to reduce to $2.5 \, \mathrm{g/L}$:

$ \mathrm{K} = \frac{1}{\mathrm{t}} \ln \frac{50}{2.5} $

Equating the two expressions for $\mathrm{K}$:

$ \frac{1}{10} \ln 2 = \frac{1}{\mathrm{t}} \ln 20 $

Solving for $\mathrm{t}$:

$ \mathrm{t} = \frac{1.3010 \times 10}{0.3010} = 43.3 \, \mathrm{sec} $

Therefore, the time required for the concentration to decrease to $2.5 \, \mathrm{g/L}$ is approximately 43 seconds.