Answer
8.5 days
Work Step by Step
Original amount $A_{0}=174\,g$
Amount after time $t$, $A=83\,g$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{8.0\,d}=0.086625\,d^{-1}$
$\ln(\frac{A_{0}}{A})=kt$
$\implies \ln(\frac{174\,g}{83\,g})=0.7402=0.086625\,d^{-1}(t)$
Then, $t=\frac{0.7402}{0.086625\,d^{-1}}=8.5\,d$
