The half-life period of a first-order chemical reaction is `6.93 min`. The time required for the completion of `99%` of the chemical reaction will be `(log 2 = 0.301)`

Options

A. 230.3 min
B. 23.03 min
C. 46.06 min
D. 460.6 min

Answer

C

Explanation

`t=(1)/(k) ln ((100)/(1))=(2ln 10 xx t_(1//2))/(ln_(2))`
`t=(2xx2.303xx6.93)/(0.693)` m in =46.06 m in