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The half life period of a first order chemical reaction is $6.93$ minutes. The time required for the completion of $99 \%$ of the chemical reaction will be $(\log 2=0.301)$ :
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The correct answer is:
$46.06$ minutes
$46.06$ minutes
$$
\begin{aligned}
& \because \lambda=\frac{0.6932}{t_{1 / 2}}=\frac{0.6932}{6.93} \min ^{-1} \\
& \text { Also } t=\frac{2.303}{\lambda} \log \frac{\left[A_{\circ}\right]}{[A]} \\
& {\left[A_{\circ}\right]=\text { initial concentration (amount) }} \\
& {[A]=\text { final concentration (amount) }} \\
& \therefore t=\frac{2.303 \times 6.93}{0.6932} \log \frac{100}{1} \\
& =46.06 \text { minutes }
\end{aligned}
$$
\begin{aligned}
& \because \lambda=\frac{0.6932}{t_{1 / 2}}=\frac{0.6932}{6.93} \min ^{-1} \\
& \text { Also } t=\frac{2.303}{\lambda} \log \frac{\left[A_{\circ}\right]}{[A]} \\
& {\left[A_{\circ}\right]=\text { initial concentration (amount) }} \\
& {[A]=\text { final concentration (amount) }} \\
& \therefore t=\frac{2.303 \times 6.93}{0.6932} \log \frac{100}{1} \\
& =46.06 \text { minutes }
\end{aligned}
$$
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