An organic compound undergoes first-order decomposition . The time taken for its decomposition 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` (take `log_(10) 2 = 0.3`)

An organic compound undergoes first-order decomposition . The time tak

1.	An organic compound undergoes first-order decomposition . The time taken for its decomposition 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` (take `log_(10) 2 = 0.3`)
Answer» Correct Answer - 9 `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`

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An organic compound undergoes first-order decomposition . The time taken for its decomposition 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` (take `log_(10) 2 = 0.3`)

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