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InterviewSolution
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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)` |
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Answer» `1^(st)` Methof : `Kt_(1//8) = ln{(C_(0))/(C_(0)//8)} = ln 8` `Kt_(1//10) = ln{(C_(0))/(C_(0)//10)} = ln 10` Then `(t_(1//8))/(t_(1//10)) xx 10 = (log 8)/(log 10) = (3 log 2)/(log 10) xx 10 = 9` `2^(nd)` Methof : `(t_(1//8))/(t_(1//10)) = (log((1)/(1//8)))/(log((1)/(1//10))) = (log 8)/(log 10) = 3 log 2 = 3 xx 0.3 = 0.9` `:. (t_(1//8))/(t_(1//10)) xx 10 = 0.9 xx 10 = 9``1^(st)` Methof : `Kt_(1//8) = ln{(C_(0))/(C_(0)//8)} = ln 8` `Kt_(1//10) = ln{(C_(0))/(C_(0)//10)} = ln 10` Then `(t_(1//8))/(t_(1//10)) xx 10 = (log 8)/(log 10) = (3 log 2)/(log 10) xx 10 = 9` `2^(nd)` Methof : `(t_(1//8))/(t_(1//10)) = (log((1)/(1//8)))/(log((1)/(1//10))) = (log 8)/(log 10) = 3 log 2 = 3 xx 0.3 = 0.9` `:. (t_(1//8))/(t_(1//10)) xx 10 = 0.9 xx 10 = 9` |
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