Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25°C. After 9 h, the fraction of sucrose remaining is f. The value of $$\log_{10}\left(\frac{1}{f}\right)$$ is ______ $$\times 10^{-2}$$.
(Rounded off to the nearest integer)
[Assume: ln10 = 2.303, ln 2 = 0.693]

For a first-order reaction, the rate constant is related to the half-life by $$k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3.33} = 0.208 \text{ h}^{-1}$$.

After time $$t = 9$$ h, the fraction of sucrose remaining is:

$$f = e^{-kt} = e^{-0.208 \times 9} = e^{-1.872}$$

We need to find $$\log_{10}\left(\frac{1}{f}\right)$$:

$$\log_{10}\left(\frac{1}{f}\right) = \log_{10}(e^{kt}) = \frac{kt}{2.303}$$

$$\log_{10}\left(\frac{1}{f}\right) = \frac{1.872}{2.303} = 0.8128$$

This equals $$81.28 \times 10^{-2}$$. Rounding to the nearest integer, the answer is $$81 \times 10^{-2}$$.

Therefore, the value is $$\textbf{81}$$.