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An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1 /$ 8 and $1 / 10$ of its initial concentration are $t_{1 / 8}$ and $t_{1 / 10}$ respectively. What is the value of $\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10$ ? $\left(\log _{10} 2=0.3\right)$
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1857 Upvotes
Verified Answer
The correct answer is:
9
$\begin{array}{l}
t_{1 / 8}=\frac{2.303 \log 8}{k}=\frac{2.303 \times 3 \log 2}{k} \\
t_{1 / 10}=\frac{2.303}{k} \log 10=\frac{2.303}{k} \\
{\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10=\frac{\left(\frac{2.303 \times 3 \log 2}{k}\right)}{\left(\frac{2.303}{k}\right)} \times 10=9}
\end{array}$
t_{1 / 8}=\frac{2.303 \log 8}{k}=\frac{2.303 \times 3 \log 2}{k} \\
t_{1 / 10}=\frac{2.303}{k} \log 10=\frac{2.303}{k} \\
{\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10=\frac{\left(\frac{2.303 \times 3 \log 2}{k}\right)}{\left(\frac{2.303}{k}\right)} \times 10=9}
\end{array}$
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