Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law , with `t_(1//2)=3.00h` . What fraction of sample of sucrose remains after 8 hours ?

Explanation

Decomposition of sucrose to glucose and fructose is a first order reaction.
`thereforek=(2.303)/(t)log.([A]_(0))/([A])`
At `t_(1//2)= 3 "hrs",k=(0.693)/(t_(1//2))=(0.693)/(3)=0.231h^(-1)`
`therefore0.231=(2.303)/(8)log.([A]_(0))/([A])`
or, `log.([A]_(0))/([A])=0.8024 " or, "([A]_(0))/([A])=6.345`
or, `([A]_(0))/([A])=(1)/(6.345)=0.158`
`therefore` The fraction that remains after 8 hours = 0.158