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//10respectively. What is the value of [t_(1//8)]/([t_(1//10)]) xx 10 (take log_(10)2=0.3)

Answer»


Solution :`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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