The half-life of a first order reaction is 100 seconds. What is the time required for 90 % completion of the reaction?
 

Half life time $\left(\frac{{\mathrm{t}}_1}{2}\right)=100\sec$. (given) $\mathrm{K}=\frac{0.693}{100}$

Time required for $90\%$ completion?
Here let initial concentration $=\mathrm{a}$
Concentration decomposed is 90 %
$\begin{array}{rl} & \mathrm{x}=\frac{90\times \mathrm{a}}{100}\\ & =0.9\mathrm{a}\end{array}$
Now,

$\begin{array}{rl}\mathrm{t}& =\frac{2.303}{\frac{0.693}{{\mathrm{t}}_1/2}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\\ & \Rightarrow \frac{2.303}{100}\log \frac{\mathrm{a}}{\mathrm{a}-0.9\mathrm{a}}\\ & \Rightarrow \frac{2.303}{0.693}\log \frac{\mathrm{a}}{0.1\mathrm{a}}\\ & \Rightarrow \frac{2.303}{100}\log 10\\ & \Rightarrow \frac{2.303}{0.693}\times 100\times \log 10\\ & \Rightarrow 339.32\sec \end{array}$