A and B decompose via first order kinetics with half-lives 54 . 0   min and 18 . 0   min…

Given $\mathrm{A}={\mathrm{t}}_{1/2}=54 \min$

$\mathrm{B}={\mathrm{t}}_{1/2}=18 \min$

$\mathrm{t}=0$ '$\mathrm{x}$' $\mathrm{M}$     $\mathrm{t}=0$ '$\mathrm{x}$' $\mathrm{M}$

$\Rightarrow$ To calculate : $\left[{\mathrm{A}}_{\mathrm{t}}\right]=16\times \left[{\mathrm{B}}_{\mathrm{t}}\right]$ time $=?$

$\Rightarrow$ For $\mathrm{I}$ order kinetic : $\left[{\mathrm{A}}_{\mathrm{t}}\right]=\frac{{\mathrm{A}}_0}{{\left(2\right)}^{\mathrm{n}}}$

$\mathrm{n}\to$ no. of Half lives

$\Rightarrow$ Now from the relation $\left(1\right)$

$\Rightarrow \frac{\mathrm{x}}{{\left(2\right)}^{{\mathrm{n}}_1}}=\frac{\mathrm{x}}{{\left(2\right)}^{{\mathrm{n}}_2}}\times 16\Rightarrow {\left(2\right)}^{{\mathrm{n}}_2}={\left(2\right)}^{{\mathrm{n}}_1}\times {\left(2\right)}^4$

$\Rightarrow {\mathrm{n}}_2={\mathrm{n}}_1+4\Rightarrow \frac{\mathrm{t}}{{\left({\mathrm{t}}_{1/2}\right)}_2}=\frac{\mathrm{t}}{{\left({\mathrm{t}}_{1/2}\right)}_1}+4$

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

$\Rightarrow \mathrm{t}=108 \min$