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Time required for \(99.9 \%\) completion of a first order reaction is _______ times the time required for completion of \(90 \%\) reaction.(nearest integer)
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$\begin{aligned} & \mathrm{K}=\frac{1}{\mathrm{t}_{99.9 \%}} \ln \left(\frac{100}{0.1}\right)=\frac{1}{\mathrm{t}_{90 \%}} \ln \left(\frac{100}{10}\right) \\ & \mathrm{t}_{99.9 \%}=\mathrm{t}_{90 \%} \frac{\ell \mathrm{n}\left(10^3\right)}{\ell \mathrm{n} 10} \\ & \mathrm{t}_{99.9 \%}=\mathrm{t}_{90 \%} \times 3\end{aligned}$
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