Half-life period of a first order reaction is 10 min. What percentage of the reaction will be completed in 100 min?

$\begin{array}{l}{\mathrm{t}}_{1/2}=10\min \mathrm{R}=\text{ ? }\\ \mathrm{K}=\frac{0.693}{{\mathrm{t}}_{1/2}}=\frac{0.693}{10}\\ \mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{R}}_{\mathrm{o}}}{\mathrm{R}}\\ \frac{0.693}{10}=\frac{2.303}{100}\log \frac{{\mathrm{R}}_{\mathrm{o}}}{{\mathrm{R}}_0-\mathrm{x}}\\ \log \frac{{\mathrm{R}}_0}{{\mathrm{R}}_0-\mathrm{x}}=\frac{0.693\times 10}{2.303}\\ \log \frac{{\mathrm{R}}_0}{{\mathrm{R}}_0-\mathrm{x}}=3.00\\ \text{ Say }{\mathrm{R}}_0-\mathrm{x}=\mathrm{R}\\ \Rightarrow \log \frac{{\mathrm{R}}_0}{\mathrm{R}}=3\Rightarrow {\mathrm{R}}_0={10}^3\mathrm{R} ....\left(1\right)\end{array}$

Now, % of reactant left after 100 min$=\frac{\mathrm{R}}{{\mathrm{R}}_0}\times 100$

$\begin{array}{l}=\frac{\mathrm{R}}{{10}^3\mathrm{R}}\times 100\left(\text{ from }\right(1\left)\right)\\ =0.1\mathrm{\%}\end{array}$

$\therefore \%$ of reaction completed

$\begin{array}{l}=100-0.1\\ =99.9\mathrm{\%}\end{array}$