For a first-order reaction, the time taken for decomposition to a certain fraction of initial concentration is given by the formula:

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

where $k$ is the rate constant, $[A{]}_0$ is the initial concentration, and $\left[A\right]$ is the concentration at time $t$.

Step 1: Find $t_{1/8}$

Here, $\left[A\right]=\frac{[A{]}_0}{8}$, so $\frac{[A{]}_0}{\left[A\right]}=8$

$t_{1/8}=\frac{2.303}{k}\times log8$

Step 2: Find $t_{1/10}$

Here, $\left[A\right]=\frac{[A{]}_0}{10}$, so $\frac{[A{]}_0}{\left[A\right]}=10$

$t_{1/10}=\frac{2.303}{k}\times log10$

Step 3: Find the ratio $\frac{t_{1/8}}{t_{1/10}}$

$\frac{t_{1/8}}{t_{1/10}}=\frac{\frac{2.303}{k}\times log8}{\frac{2.303}{k}\times log10}=\frac{log8}{log10}$

Since $log10=1$, we get:

$\frac{t_{1/8}}{t_{1/10}}=log8=log{2}^3=3\times log2\approx 3\times 0.3010=0.903$

Step 4: Calculate the required expression

We need to find $\frac{t_{1/8}}{t_{1/10}}\times 10$

Substituting the value from above:

$\frac{t_{1/8}}{t_{1/10}}\times 10=0.903\times 10=9.03$

Final Answer: The value is approximately 9.03.

Related Topics and Formulae

First-Order Kinetics: The rate of reaction depends linearly on the concentration of one reactant. The integrated rate law is $k=\frac{2.303}{t}log\frac{[A{]}_0}{\left[A\right]}$.

Half-Life: For a first-order reaction, half-life $t_{1/2}=\frac{0.693}{k}$ is constant and independent of initial concentration.

Fractional Life: The time for decomposition to any fraction $\frac{1}{n}$ is proportional to $logn$.