Question

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 $\frac{\left[t_{1/8}\right]}{\left[t_{\nu 70}\right]}\times 10$ ? (take ${\log }_{10}2=$ $0.3$ )

Answer 1

Answer: 9

$l_{1/8}=\frac{2.303\log 8}{k}=\frac{2.303\times 3\log 2}{k}$
$t_{1/10}=\frac{2.303}{k}\log 10=\frac{2.303}{k}$
$\left[\frac{t_{18}}{t_{1/10}}\right]\times 10=\frac{\left(\frac{2.303\times 3\log 2}{k}\right)}{\left(\frac{2.303}{k}\right)}\times 10=9$