Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $$t_{1/2} = 3$$ hour. The percentage of sucrose remaining after 6 hours is _______. (Nearest integer)
(Given: log 2 = 0.3010 and log 3 = 0.4771)

For a first-order reaction, the half-life remains constant.

The number of half-lives elapsed in $$6\ \text{hours}$$ is

$$n=\frac{t}{t_{1/2}}=\frac{6}{3}=2$$

After the first half-life:

$$100%\rightarrow 50%$$

After the second half-life:

$$50%\rightarrow 25%$$

Alternatively,

$$\text{Amount remaining}=\frac{\text{Initial amount}}{2^n}=\frac{100}{2^2}=25%$$

Therefore, the percentage of sucrose remaining after $$6\ \text{hours}$$ is

$$25%$$

Hence, the correct answer is $$25$$.