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)`

Explanation

`1^(st)` Methof :
`Kt_(1//8) = ln{(C_(0))/(C_(0)//8)} = ln 8`
`Kt_(1//10) = ln{(C_(0))/(C_(0)//10)} = ln 10`
Then `(t_(1//8))/(t_(1//10)) xx 10 = (log 8)/(log 10) = (3 log 2)/(log 10) xx 10 = 9`
`2^(nd)` Methof :
`(t_(1//8))/(t_(1//10)) = (log((1)/(1//8)))/(log((1)/(1//10))) = (log 8)/(log 10) = 3 log 2 = 3 xx 0.3 = 0.9`
`:. (t_(1//8))/(t_(1//10)) xx 10 = 0.9 xx 10 = 9`