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

Explanation

For the first order reaction: `k-0.693/(t_(1//2)) = 0.693/(3.0 hr)`
`t=2.303/k log [A]_(0)/[A] = ( k xx t)/(2.303)`
`log [A]_(0)/[A] = 0.693/(3 hr) xx (8hr)/2.303 = 0.8024`
`log [A]_(0)/[A] = 0.693/(3 hr) xx (8 hr)/(2.303) = 0.8024`
`[A]_(0)/[A]` = Antilog `0.8024 = 6.3445`
`[A]/[A]_(0) = 1/6.3445 = 0.158`