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)`

Answer

`(B) rarr 8`

Explanation

For a first order reaction
`t=(2.303)/(t)"log"(c_(0))/(c_(t))ort=(2.303)/(k)"log"(c_(0))/(c_(t))`
`:. t_(1//8)=(2.303)/(k)"log"(c_(0))/(18)=(2.303)/(k)log`
`=(2.303)/(k)"log"^(2^(3))=(2.303)/(k)log_(2)`
`t_(1//10)=(2.303)/(k)"log"(c_(0))/(c_(0)//10)=(2.303)/(k)"log"_(10)=(2.303)/(k)`
`:. =(t_(1//8))/(t_(1//10))=3xx0.3=0.9or (t_(1//8))/(t_(1//10))xx10=9`.