Number of natural life times `(T_(av))` required for a first order reaction to achieve `99.9%` level of comletion is

Options

A. `2.3`
B. `6.9`
C. `9.2`
D. `0.105`

Answer

B

Explanation

If no. of half lives = n
`[A] = ([A]_(0))/(e^(n))`
Here `[A]_(0)-[A]=99.9%` of `[A]_(0)`
`[A]=((100-99.9)/(100))[A]_(0)=10^(-3)[A]_(0)`
`therefore 10^(-3)[A]_(0)=([A]_(0))/(e^(n))`
`e^(n)=10^(3)`
`log_(e )e^(n)=log_(e )10^(3)`
`n=3xx2.303 log_(10)10`
`=6.909xx1=6.909`