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An organic compound undergoes first-orde...

An organic compound undergoes first-order decomposition. The time taken for this 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 2 = 0.3)

Text Solution

Verified by Experts

The correct Answer is:
9
Using the expression for first-order reaction, `t = (2.303)/(k)log""([A_(0)])/([A])`
when the compound is decomposed to 1/8 th of its initial value then the time taken is
`t_(1//8) = ((2.303)/(k))log""(1)/((1//8)) = ((2.303)/(k))log 8` …..(1)
when the compound is decomposed to 1/10 th of its initial value then the time taken is
`t_(1//10) = ((2.303)/(k))log""(1)/((1//10)) = ((2.303)/(k)) log10` ...(2)
Dividing Eq. (1) by Eq. (2), we get, `(t_(1//8))/(t_(1//10)) = (log 8)/(log 10) = log (2)^(3) = 3 xx 0.3 = 0.9`
So, the value of `([t_(1//8)])/([t_(1//10)]) xx 10 = 9`.
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