therefore, Fall of concentration from $1.28 \;\mathrm{mg}\, \mathrm{L}^{-1}$ to $0.04\; \mathrm{mg} \,\mathrm{L}^{-1}$ involves five half-lives.

$1.28 \stackrel{t_{1 / 2}}{\rightarrow} 0.64 \stackrel{t_{1 / 2}}{\rightarrow} 0.320 .16 \stackrel{t_{1 / 2}}{\rightarrow} 0.08 \stackrel{t_{1 / 2}}{\rightarrow} 0.04$

therefore, Time required $=5 \times t_{1 / 2}$

$=5 \times 138\; \mathrm{s}$

$=690\; \mathrm{s}$