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

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