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 )

Chemistry · Chemical Kinetics

JEE Main 2026 — 6 April, Morning Shift — Question 66

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $\mathrm{t}_{1 / 2}=3$ hour. The percentage of sucrose remaining after 6 hours is $\_\_\_\_$ . (Nearest integer) (Given : $\log 2=0.3010$ and $\log 3=0.4771$ )

Answer: 25

Numerical answer — enter this value.

Step-by-step solution

Sucrose $+\mathrm{H}_{2} \mathrm{O} \rightarrow$ glucose + fructose $\mathrm{t}_{1 / 2}=3 \mathrm{hr}$ For $1^{\text {st }}$ order

\mathrm{t}_{75 \%}=2 \times \mathrm{t}_{1 / 2}

So $\mathrm{t}=6 \mathrm{hr}$ , 75% of sucrose decomposed So $25 \%$ of sucrose remains after 6 hr .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws