For a first order reaction, the amount of reactant remaining after n half-lives is given by Nt=2nN0.
Given t1/2=3 hours and total time t=6 hours.
Number of half-lives, n=t1/2t=36=2.
The amount of sucrose remaining is Nt=22N0=4N0.
Percentage of sucrose remaining =N0Nt×100=41×100=25.
Answer: 25