First order reaction,

k = \(\frac{2.303}{t}\)log \(\frac{[R]_0}{[R]}\)

It is given that, t_1/2 = 3.00 hours

k = \(\frac{0.693}{t_{1/2}}\)

Therefore, \(\frac{0.693}{3}h^{-1}\)

= 0.231 h^-1

Then, 0.231 h^-1

= \(\frac{2.303}{8 \ h}\)log \(\frac{[R]_0}{[R]}\)

⇒ log \(\frac{[R]_0}{[R]}\) = \(\frac{0.231 h^{-1} \times 8 h}{2.303}\)

⇒ \(\frac{[R]_0}{[R]}\) = antilog (0.8024)

⇒ \(\frac{[R]_0}{[R]}\) = 6.3445

⇒ \(\frac{[R]_0}{[R]}\) = 0.1576 (approx)

= 0.158

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.