The decomposition of a compound follows the rate law of a first order reaction. For the initial concentration of the compound to drop to `(1)/(8)` and `(1)/(10)` th of this value , times required are `t_(1//8)th and t_(1//10)` respectively.
Find the value of `(t_(1//8)/(t_(1//10))xx10). ["given":log"10^(2)=0.3]`

Explanation

The rate constant of a first order reaction,
`k=(2.303)/(t)log.([A]_(0))/([A])`
Given , for `t=t_(1//8),[A]=(1)/(8)[A]_(0)`
and for `t=t_(1//10),[A]=(1)/(10)[A]_(0)`
`thereforet_(1//8)=(2.303)/(k)log.([A]_(0))/([A])=(2.303)/(k)log.([A]_(0))/((1)/(8)[A]_(0))`
`=(2.303)/(k)log8=(3xx2.303)/(k)log2=(3xx2.303xx0.3)/(k)`
and `t_(1//10)=(2.303)/(k)log.([A]_(0))/([A])=(2.303)/(k)log.([A]_(0))/((1)/(10)[A]_(0))=(2.303)/(k)`
`therefore(t_(1//8))/(t_(1//10))xx10=((3xx2.303xx0.3)/(k))/((2.303)/(k))xx10=9`