Initial concentration of reactant in a first order reaction is \(0.08 \mathrm{~mol} \mathrm{dm}^{-3}\) What concentration would remain after 40 minute?
\(\left(\operatorname{given} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=5.00\right)\)

For a first order reaction,
\(\begin{aligned}
\mathrm{t} & =\frac{2.303}{\mathrm{k}} \log _{10} \frac{[\mathrm{~A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\therefore \quad \mathrm{k} & =\frac{2.303}{40 \text { minute }} \log _{10} 5=0.04
\end{aligned}\)
Now,
\(\begin{array}{ll}
\therefore & 0.04=\frac{2.303}{40} \log _{10} \frac{0.08}{[\mathrm{~A}]_{\mathrm{t}}} \\
\therefore & 0.69=\log _{10}(0.08)-\log _{10}[\mathrm{~A}]_{\mathrm{t}} \\
\therefore & \log _{10}[\mathrm{~A}]_{\mathrm{t}}=-1.096-0.69 \\
\therefore & \log _{10}[\mathrm{~A}]_{\mathrm{t}}=-1.78 \\
\therefore & {[\mathrm{~A}]_{\mathrm{t}}=\operatorname{Antilog}_{10}(-1.78)=0.016}
\end{array}\)
Alternate method:
\(\begin{aligned}
& \text { Given } \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=5 \\
\therefore \quad & {[\mathrm{~A}]_{\mathrm{t}}=\frac{0.08}{5}=0.016 }
\end{aligned}\)