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If half-life of a substance is 36 minutes. Find amount left after 2 hrs . Initial amount is 10 gm ?
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Verified Answer
The correct answer is:
1 g
Given,
$\mathrm{t}_{1 / 2}=36 \mathrm{~min}, \mathrm{t}=2 \mathrm{hrs}=120 \mathrm{~min},[\mathrm{Ao}]=\log$
For first order reaction, $\mathrm{R}=\frac{0.693}{\mathrm{t}_{1 / 2}}$
$\begin{aligned}
\therefore \quad k & =\frac{0.693}{36}=0.01925 \\
t & =\frac{2303}{k} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}
\end{aligned}$
where, $\left[\mathrm{A}_0\right]=$ initial amount
$\begin{aligned}
& {[\mathrm{A}]=\text { final amount } } \\
& 120=\frac{2303}{0.01925} \log \frac{10}{[\mathrm{~A}]} \\
& \log 10-\log [\mathrm{A}]=1 \\
& 1-\log [\mathrm{A}]=1 \\
& \log [\mathrm{A}]=0 \\
& {[\mathrm{~A}] }=1 \mathrm{gm}
\end{aligned}$
$\mathrm{t}_{1 / 2}=36 \mathrm{~min}, \mathrm{t}=2 \mathrm{hrs}=120 \mathrm{~min},[\mathrm{Ao}]=\log$
For first order reaction, $\mathrm{R}=\frac{0.693}{\mathrm{t}_{1 / 2}}$
$\begin{aligned}
\therefore \quad k & =\frac{0.693}{36}=0.01925 \\
t & =\frac{2303}{k} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}
\end{aligned}$
where, $\left[\mathrm{A}_0\right]=$ initial amount
$\begin{aligned}
& {[\mathrm{A}]=\text { final amount } } \\
& 120=\frac{2303}{0.01925} \log \frac{10}{[\mathrm{~A}]} \\
& \log 10-\log [\mathrm{A}]=1 \\
& 1-\log [\mathrm{A}]=1 \\
& \log [\mathrm{A}]=0 \\
& {[\mathrm{~A}] }=1 \mathrm{gm}
\end{aligned}$
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