Answer
8.5 days
Work Step by Step
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{8.0\,d}=0.086625\,d^{-1}$
Original amount $A_{0}=174\,g$
The amount remaining $A=83\,g$
Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required.
$\implies \ln(\frac{174\,g}{83\,g})=0.7402=0.086625\,d^{-1}(t)$
$\implies t=\frac{0.7402}{0.086625\,d^{-1}}=8.5\,d$
