\(A\) \(\quad\) \(B\)

\(\Rightarrow \quad\) To calculate \(:\left[ A _{t}\right]=16 \times\left[ B _{ t }\right] \ldots .(1)\) time \(=?\)

\(\Rightarrow \quad\) For \(I\) order kinetic : \(\left[ A _{ t }\right]=\frac{ A _{0}}{(2)^{ n }}\)

\(n \rightarrow\) no of Half lives

\(\Rightarrow \quad\) Now from the relation \((1)\)

\(\left[ A _{ t }\right]=16 \times\left[ B _{ t }\right]\)

\(\Rightarrow \quad \frac{x}{(2)^{n_{1}}}=\frac{x}{(2)^{n_{2}}} \times 16 \Rightarrow(2)^{ n _{2}}=(2)^{n_{1}} \times(2)^{4}\)

\(\Rightarrow \quad n_{2}=n_{1}+4 \quad \Rightarrow \quad \frac{t}{\left(t_{1 / 2}\right)_{2}}=\frac{t}{\left(t_{1 / 2}\right)_{1}}+4\)

\(\Rightarrow \quad t\left(\frac{1}{18}-\frac{1}{54}\right)=4 \Rightarrow t=\frac{4 \times 18 \times 54}{36}\)

\(\Rightarrow \quad t=108\, \min\)