Question
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: 2 = 0.3010 and 3 = 0.4771)
JEE Main
Chemistry
Biomolecules
2026
JEE Main 2026 (06 April Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
For a first order reaction, the amount of reactant remaining after n half-lives is given by N_t = N_0 2^n . Given t_ 1/2 = 3 hours and total time t = 6 hours. Number of half-lives, n = t t_ 1/2 = 6 3 = 2. The amount of sucrose remaining is N_t = N_0 2^2 = N_0 4 . Percentage of sucrose remaining = N_t N_0 100 = 1 4 100 = 25. Answer: 25
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