An organic compound undergoes first decompoistion. The time taken for its decompoistion to `1//8` and `1//10` of its initial concentration are `t_(1//8)` and `t_(1//10)`, respectively. What is the value of `([t_(1//8)])/([t_(1//10)]) xx 10`? `(log_(10)2 = 0.3)`

Answer

9

Explanation

`K=2.303/t"log"a_(0)/a`
`:' a=a_(0)/8 at t_(1//8)`
`:. T_(1//8)=2.303/K"log"a_(0)/(a_(0)//8)=2.303/K"log" 8 …(1)`
`:' a=a_(0)//10 at t_(1//10)`
`t_(1//10)=2.303/K"log"a_(0)/(a_(0)//10)=2.303/K"log" 10 …(2)`
From eq. (1) and (2)
`([t_(1//8)])/([t_(1//10)])xx10=2.303/K"log" 8xx(K)/(2.303xxlog 10)xx10`
`=log8/log10xx10=log2^(3)/log10xx10=(3log2)/(log10)xx10`
`=(3xx0.3xx10)/(1)=9`