Given data

• First-order hydrolysis
• Half-life: $t_{1/2} = 3.33\,\text{h}$
• Time elapsed: $t = 9\,\text{h}$
• Fraction of sucrose remaining: $f$

First-order relation for half-life For a first-order reaction, $k = \frac{0.693}{t_{1/2}}$ So, $k = \frac{0.693}{3.33} = 0.2081\,\text{h}^{-1} \; (\text{approx})$

Fraction remaining after time $t$ For a first-order reaction, $f = e^{-kt}$ Hence, $\frac{1}{f} = e^{kt}$ Taking common logarithm, $\log_{10}\left(\frac{1}{f}\right) = \frac{kt}{2.303}$

Substitute values $\log_{10}\left(\frac{1}{f}\right) = \frac{(0.2081)(9)}{2.303}$ $= \frac{1.8729}{2.303}$ $\approx 0.813$

Convert into the asked form We need $\log_{10}\left(\frac{1}{f}\right) = \underline{\hspace{1cm}} \times 10^{-2}$ Since $0.813 = 81.3 \times 10^{-2}$ Rounded to nearest integer, $\boxed{81}$

Comparison with stored answer Derived answer = $81$

Stored correct answer = $81$

Hence, they agree.