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

Given : `t_(1//2)=3` hours, t=8 hours
As sucrose decomposes according to first order rate law,
`k=(2*303)/(t_(1//2))"log"([A]_(0))/([A]_(t))`
Using the formula `k=(0*693)/(t_(1//2))`
`:." "k=(0*693)/(t_(1//2))`
`=(0*693)/(3)=0*231hr^(-1)`
From formula, `k=(2*303)/(t)"log"([A]_(0))/([A]_(t))`
`:." "0*231=(2*303)/(8)"log"([A]_(0))/([A]_(t))`
`:." log"([A]_(0))/([A]_(t))=0.8024`
Taking antilog on both sides,
`:." "([A]_(0))/([A]_(t))=" Antilog "(0*8024)`
`=6*345`
`:." "([A]_(0))/([A]_(t))=(1)/(6*345)=*158`
Hence, the fraction of the sample of sucrose that remains after 8 hours is `0*158`.