Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law,
with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

$$\begin{gathered} {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\mathrm{k}} \\ \mathrm{k}=\frac{0.693}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \mathrm{k}=\frac{0.693}{3}=0.231 \end{gathered}$$

Let initial concentration of sucrose be 1 M.
After 8 hours, the concentration will be 1−x. 

x represents the amount of sucrose decomposed.
$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\left[\mathrm{A}\right]}_{\mathrm{o}}}{\left[\mathrm{A}\right]}$

$$\begin{gathered} ⟹0.231= \frac{2.303}{8} \log \left(\frac{1}{1-\mathrm{x}}\right) \\ ⟹\log \left(\frac{1}{1-\mathrm{x}}\right)=0.8024 \\ ⟹\frac{1}{1-\mathrm{x}}=6.345 ⟹\mathrm{x}=0.842 \end{gathered}$$

After 8 hours, the concentration of sucrose left = 1−0.842=0.158 M.