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

An organic compound undergoes first order decomposition. The time take

1.	An organic compound undergoes first order decomposition. The time taken for the decomposition of `1//8` and `1//10` of its initial concentration are `1//8` and `1//10` respectively. What is the value of `[t_(1//8)]/([t_(1//10)]) xx 10` (take `log_(10)2=0.3)`
Answer» `k=2.303/t log ([A]_(0))/([A])` `[A] = 1//8 [A]_(0)` at `t_(1//8)` `t_(1//8) = 2.303/k log ([A]_(0))/([A]_(0//8))` `=2.303/k log8` `[A] = 1/10 [A]_(0)` at `t_(1//10)`, `t_(1//10) = 2.303/k log ([A]_(0))/([A]_(0//10))` `=2.303/k log 10`...........(ii) `([t_(1//8)])/([t_(1//10)]) xx 10 = (log 8)/(log 10) xx 10 = (3 log 2)/(log 10) xx 10` `= 3 xx (0.3010 xx 10) =9`

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

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