Q24 / 75 · JEE Main 2026 (06 April Shift 1) · Chemistry
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 $\log 3 = 0.4771$)
Correct answer: 25
Solution
For a first order reaction, the amount of reactant remaining after $n$ half-lives is given by $N_t = \dfrac{N_0}{2^n}$.
Given $t_{1/2} = 3$ hours and total time $t = 6$ hours.
Number of half-lives, $n = \dfrac{t}{t_{1/2}} = \dfrac{6}{3} = 2$.
The amount of sucrose remaining is $N_t = \dfrac{N_0}{2^2} = \dfrac{N_0}{4}$.
Percentage of sucrose remaining $= \dfrac{N_t}{N_0} \times 100 = \dfrac{1}{4} \times 100 = 25$.
Answer: $25$
Given $t_{1/2} = 3$ hours and total time $t = 6$ hours.
Number of half-lives, $n = \dfrac{t}{t_{1/2}} = \dfrac{6}{3} = 2$.
The amount of sucrose remaining is $N_t = \dfrac{N_0}{2^2} = \dfrac{N_0}{4}$.
Percentage of sucrose remaining $= \dfrac{N_t}{N_0} \times 100 = \dfrac{1}{4} \times 100 = 25$.
Answer: $25$
Hence, the correct answer is 25.