Sol. $$ \begin{aligned} & t_{10 \%}=\frac{1}{K} \ln \left(\frac{a}{a-x}\right)=\frac{1}{K} \ln \left(\frac{100}{90}\right) \\ & t_{10 \%}=\frac{2.303}{K}(\log 10-\log 9) \\ & t_{10 \%}=\frac{2.303}{K} \times(0.04) \end{aligned} $$ Similarly $\begin{aligned} & t_{90 \%}=\frac{1}{K} \ln \left(\frac{100}{10}\right) \\ & t_{90 \%}=\frac{2.303}{K} \\ & \frac{t_{90 \%}}{t_{10 \%}}=\frac{1}{0.04}=25 \\ & e^{k t}=\frac{a}{a-x} \\ & \frac{a-x}{a}=e^{-k t} \\ & 1-\frac{x}{a}=e^{-k t} \\ & x=a\left(1-e^{-k t}\right) \\ & \alpha=\frac{x}{a}=\left(1-e^{-k t}\right)\end{aligned}$