For a first order reaction, the ratio of time for the completion of `99.9%` and half of the reaction is

Options

A. 2
B. 4
C. 10
D. 8

Answer

C

Explanation

`t_(1//2)(2.303)/(k)log2`
`t_(99.9%)=(2.303)/(k)"log"(100)/(100-99.9)=(2.303)/(k)"log"(100)/(0.1)`
`=(2.303)/(k)log10^(3)=(3xx2.303)/(k)=(3xxt_(1//2))/(log2)`
`=(3xxt_(1//2))/(0.3010)~~10t_(1//2)`.