\({C_o}\, - \,{C_t}\, = \,Kt\)
\(0.5M\, - \,0.2M\, = \,Kt\)
\(0.3\, = \,Kt\) ....... \((1)\)
\('K'\) can be calculated by
\({t_{1/2}}\, = \,\frac{{{C_o}}}{{2K}}\)
\(6\, = \,\frac{{0.2}}{{2K}}\)
\(K\, = \,\frac{{0.2}}{{12}}\, = \,\frac{{2 \times {{10}^{ - 1}}}}{{12}}\, = \,\frac{1}{{60}}\)
Putting the value of \(K\) in eq \((1)\)
\(t\, = \,\frac{{0.3}}{K}\, = \,\frac{{0.3}}{{1/60}}\, = \,60\, \times \,0.3\, = \,18\,Hr\)
(આપેલ : $\ln 10=2.303\,\log 2=0.3010$ )