20% first order reaction is completed in 50 minutes. Time required for the completion of 60% of the reaction is.......min

                         $K_{20\%}=K_{60\%}$

    $\frac{2.303}{50}\log \frac{100}{80}=\frac{2.303}{t}\log \frac{100}{40}$

     $\frac{\log 10-\log 8}{50}=\frac{\log 10-\log 4}{t}$

             $\frac{1-0.9030}{50}=\frac{1-0.6020}{t}$

                               $t=205\min$