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

9

Explanation

For a first order reaction,
`k=(2*303)/t"log"C_0/C_t" or "t=(2*303)/k"log"C_0/C_t`
`:." "1_(1//8)=(2*303)/t"log"C_0/(C_0//8)=(2*303)/k"log 8"`
`=(2*303)/k"log 2"^3=(2*303xx3)/k"log 2"`
`=(2*303xx3xx0*3)/k`
`t_(1//2)=(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*9" or "t_(1//8)/(t_(1//10))" or "t_(1//8)/t_(1//10)xx10=9`