The ratio of the time taken for `99.9%` reaction and half life of the reaction is :

Options

A. 100
B. 49.95
C. 10
D. 1000

Answer

C

Explanation

( C) For 99.9% completion
`R = [R]_(0)-(99.9)/(100) [R]_(0) = 0.001 [R]_(0)`
` t(99.9%) = (2.303) /(k) "log " ([R]_(0))/(0.001[R]_(0))`
`= (2.303)/(k) "log" 10^(3)`
`= (2.303xx3)/(k)`
`t_(1//2) = (2.303)/(k) "log" ([R]_(0))/(0.5[R]_(0)) `
`= (2.303"log"2)/(k)`
`=(2.303xx0.3010)/(k)`
Dividing eq . (i ) by eq (ii) we get
`(t(99.9%))/(t_(1//2))= (3)/(0.3010) ~~ 10`