c
$k = {\mkern 1mu} {\mkern 1mu} \frac{{0.693}}{{{t_{1/2}}}}{\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} \frac{{0.693}}{{69.3}}\,\min$ ${\mkern 1mu} (\because {\mkern 1mu} {\mkern 1mu} \,{t_{1/2}}{\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} 69.3{\mkern 1mu} {\mkern 1mu} \,\min .)$
હવે , $k = \,\,\frac{{2.303}}{t}\,\,\log \,\,\frac{{100}}{{20}}\,\,\,\,\,\,\,\,[\,\,a\,\, = \,\,100,\,\,x\,\, = \,\,80,\,\,\,a\, - \,\,x\,\, = \,\,20]$
$\therefore \,\,\frac{{0.693}}{{69.3}}\,\, = \,\,\frac{{2.303}}{t}\,\,\log \,\,5\,\,\,\,\,\,\therefore \,\,t\,\, = \,\,160.97\,\min$