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 $\log 3=0.4771$ )
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Solution
Sucrose $+\mathrm{H}_2 \mathrm{O} \rightarrow$ glucose + fructose
$$
t_{1 / 2}=3 \mathrm{hr}
$$
For $1^{\text {st }}$ order $\quad t_{75 \%}=2 \times t_{1 / 2}$
So $t=6 \mathrm{hr}, \quad 75 \%$ of sucrose decomposed
So $25 \%$ of sucrose remains after 6 hr
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