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

`Kt_(1//8) =` ln `{(C_(O))/(C_(O)//8)} = `ln 8
`K t_(1//10)` = ln `{(C_(O))/(C_(O)//10)}` = ln 10
then `(t_(1//8))/(t_(1//10)) xx 10 = ("ln"8)/("ln"10) xx 10 = (3"log"2)/("log"10) xx 10 = 9`