Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25$^\circ$C. After 9 h, the fraction of sucrose remaining is f. The value of ${\log _{10}}\left( {{1 \over f}} \right)$ is ________ $\times$ 10^$-$2. (Rounded off to the nearest integer)

[Assume : ln 10 = 2.303, ln 2 = 0.693]

Given, $$\mathop {{C_{12}}{H_{22}}{O_{11}}}\limits_{Sucrose} + {H_2}O\buildrel {1st\,order} \over \longrightarrow \mathop {{C_6}{H_{12}}{O_6}}\limits_{Glu\cos e} + \mathop {{C_6}{H_{12}}{O_6}}\limits_{Fructose} $$<br/><br/>${t_{1/2}} = {{10} \over 3}h$<br/><br/>At t = 0, a = [A]₀ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(initial conc.)<br/><br/>t = 9h, a $-$ x = [A]_t &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[conc. at time t]<br/><br/>For using 1st order equation,<br/><br/>$K = {{2.303} \over t}\log {{{{[A]}_0}} \over {{{[A]}_t}}}$ <br/><br/>$\Rightarrow {{K \times t} \over {2.303}} = \log {{{{[A]}_0}} \over {{{[A]}_t}}}$<br/><br/>$${{\ln 2 \times 9} \over {10/3 \times 2.303}} = \log \left( {{1 \over F}} \right) \Rightarrow \log \left( {{1 \over F}} \right) = 0.8124$$ ($\because$ $k = {{\ln 2} \over {{t_{1/2}}}}$)<br/><br/>$\log \left( {{1 \over F}} \right) = 81.24 \times {10^{ - 2}}$<br/><br/>x = 81.24 or x $\approx$ 81