For the reaction $$A \to B$$ the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to 2.5 g L$$^{-1}$$ (if the initial concentration of A was 50 g L$$^{-1}$$) is _________.
(Nearest integer)
Given : $$\log 2 = 0.3010$$

By analyzing the given concentration versus time graph, we observe that:

• At $$t = 0,\text{s}$$, the initial concentration is $$[A]_0 = 50,\text{g L}^{-1}.$$
• At $$t = 10,\text{s}$$, the concentration becomes $$[A] = 25,\text{g L}^{-1}.$$

Since the concentration decreases from $$50,\text{g L}^{-1}$$ to $$25,\text{g L}^{-1}$$ in $$10,\text{s}$$, the half-life of the reaction is

$$t_{1/2} = 10,\text{s}.$$

A constant half-life is characteristic of a first-order reaction.

For a first-order reaction,

$$t=\frac{2.303}{k}\log\left(\frac{[A]_0}{[A]}\right).$$

The relationship between the rate constant and half-life is

$$k=\frac{2.303\log 2}{t_{1/2}}.$$

Substituting this expression for $$k$$ into the integrated rate equation,

$$t=\frac{t_{1/2}}{\log 2}\log\left(\frac{[A]_0}{[A]}\right).$$

Using the given data,

• $$t_{1/2}=10,\text{s}$$
• $$[A]_0=50,\text{g L}^{-1}$$
• $$[A]=2.5,\text{g L}^{-1}$$
• $$\log 2=0.3010$$

we get

$$t=\frac{10}{0.3010}\log\left(\frac{50}{2.5}\right).$$

Since

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

the equation becomes

$$t=\frac{10}{0.3010}\log(20).$$

Using the logarithmic identity,

$$\log(20)=\log(2\times10)=\log2+\log10=0.3010+1=1.3010,$$

we obtain

$$t=\frac{10}{0.3010}\times1.3010,$$

$$t=\frac{13.010}{0.3010}\approx43.22,\text{s}.$$

Rounding to the nearest integer,

$$\boxed{43}$$

Hence, the required time is

$$\boxed{43\ \text{s}}.$$