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 )

Step-by-step solution

For a first order reaction, the amount of reactant remaining after n half-lives is given by N_t = N₀ 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 = N₀ 2^2 = N₀ 4 . Percentage of sucrose remaining = N_t N₀ 100 = 1 4 100 = 25 . Answer: 25