In a first order reaction, 20 % reaction is completed in 24 minutes. The percentage of reactant remaining after 48 minutes is

For a first order reaction:

$$\begin{gathered} \mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\ Case \left(1\right) : \mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{100}{100-60} \\ \begin{array}{rl} & =\frac{2.303}{20}\log \frac{100}{40}\\ & =\frac{2.303}{20}(\log 10-\log 4)\\ & =\frac{2.303}{20}(1-0.6020)\\ & =0.046\end{array} \\ Case \left(\mathrm{ii}\right) : \mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{100}{100-84} \\ \begin{array}{rl} & \mathrm{t}=\frac{2.303}{0.046}\log \frac{100}{16}\\ & =\frac{2.303}{0.046}(\log 100-\log 16)\\ & =\frac{2.303}{0.046}(2-1.2041)\\ & =39.84\min =40\min \end{array} \end{gathered}$$