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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

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 \(\log 3 = 0.4771\))

  1. A 10
  2. B 15
  3. C 20
  4. D 25
Verified Solution

Answer & Solution

Correct Answer

(D) 25

Step-by-step Solution

Detailed explanation

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…
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