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

As sucrose decomposes according to first order rate law, `k=(2.303)/(t)log""([A]_(0))/([A])`
The aim is to find `[A]//[A]_(0)`
As `t_(1//2)=3.0" hour,":.k=(0.693)/(t_(1//2))=(0.693)/(3" hr")=0.231" hr"^(-1)`
Hence, `0.231" hr"^(-1)=(2.303)/(8" hr")log""([A]_(0))/([A])" or "log""([A]_(0))/([A])=0.8024" or "([A]_(0))/([A])=" Antilog "(0.8024)=6.345`
or `([A])/([A]_(0))=(1)/(6.345)=0.158.`