C12H22O11+H2O→t1/2=310hrorderGlucoseC6H12O6+FructoseC6H12O6
t=0a=[A]0
t=9hra−x=[A]t
from I order kinetic: 2.303k×t=log∣A∣t∣A∣0
⇒310×2.303ln2×9=log(f1)
⇒23.030.693×9×3=log(f1)
⇒log(f1)=0.81246=81.24×10−2
⇒x=81
JEE Main 2021 — Chemistry Physical Chemistry
Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33h at 25∘C. After 9h, the fraction of sucrose remaining is f. The value of log10(f1) is ________ ×10−2. (Rounded off to the nearest integer)
[Assume: In10=2.303,ln2=0.693]
Held on 24 Feb 2021 · Verified 6 Jul 2026.
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