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)`

Options

A. 2
B. 3
C. 3
D. 9

Answer

D

Explanation

For a first order process, `kt="In[A]_(0)/([A])`
`rArrkt_(1//8)=In[A]_(0)/([A]_(0)//8)="In "8`
`and" "kt_(1//10)="In"[A]_(0)/([A]_(0)//10)="In "10` …..(ii)
Therefore, "
`(t_(1//8))/(t_(1//8))= ("In 8")/("In 10")=log8 = 3 log 2 = 3 xx 0.3= 0.9 `
` rArr " "t_(1//8)/t_(1//10)xx 10 xx 0.9 xx 10 = 9.0 `