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( 1 / f ) is ________ x 10^(-2). (Rounded off to the nearest integer)[Assume : ln 10 = 2.303, ln 2 = 0.693]

K=frac0.693t_1 / 2 => (0.693 / 3.33) => 0.208 K=(2.303 / t) log (R_0 / R) 0.208=(2.303 / 9) log (1 / f) log (1 / f)=0.813=81.3 x 10^-2

Sucrose hydrolyses in acid solution into glucose and fr... | (Chemistry) | DivineJEE

Exam

Chemistry

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]

$K=\frac{0.693}{t_{1 / 2}} \Rightarrow \frac{0.693}{3.33} \Rightarrow 0.208 \\$

$K=\frac{2.303}{t} \log \frac{R_0}{R} \\$

$0.208=\frac{2.303}{9} \log \frac{1}{f} \\$

$\log \frac{1}{f}=0.813=81.3 \times 10^{-2}$

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