An organic compound undergoes first order decomposition. The time taken for decomposition to $$\left(\frac{1}{8}\right)^{th}$$ and $$\left(\frac{1}{10}\right)^{th}$$ of its initial concentration are $$t_{1/8}$$ and $$t_{1/10}$$ respectively. What is the value of $$\frac{t_{1/8}}{t_{1/10}}\times 10$$ ?
$$(\log{2}=0.3)$$

We need to find the ratio $$\frac{t_{1/8}}{t_{1/10}} \times 10$$ for a first-order decomposition reaction.

For a first-order reaction, the time taken for the concentration to decrease from $$C_0$$ to $$C$$ is given by

$$t = \frac{2.303}{k} \log_{10}\left(\frac{C_0}{C}\right)$$

where $$k$$ is the rate constant.

To obtain $$t_{1/8}$$, the time required for the concentration to reach $$\frac{C_0}{8}$$, we substitute $$C = \frac{C_0}{8}$$ to get

$$t_{1/8} = \frac{2.303}{k} \log_{10}\left(\frac{C_0}{C_0/8}\right) = \frac{2.303}{k} \log_{10}(8)$$

Since $$8 = 2^3$$, it follows that

$$t_{1/8} = \frac{2.303}{k} \times 3\log_{10}(2) = \frac{2.303 \times 3 \times 0.3}{k} = \frac{2.303 \times 0.9}{k}$$

Similarly, for $$t_{1/10}$$, corresponding to $$C = \frac{C_0}{10}$$, we have

$$t_{1/10} = \frac{2.303}{k} \log_{10}\left(\frac{C_0}{C_0/10}\right) = \frac{2.303}{k} \log_{10}(10) = \frac{2.303}{k} \times 1 = \frac{2.303}{k}$$

Dividing these results yields

$$\frac{t_{1/8}}{t_{1/10}} = \frac{\frac{2.303 \times 0.9}{k}}{\frac{2.303}{k}} = 0.9$$

Finally, multiplying by 10 gives

$$\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$$

The correct answer is Option 3: 9.