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. 22.03 min
C. 46.06 min
D. 460.6 min

Answer

C

Explanation

`k_(1)=(0.693)/(6.93min)=0.1"min"^(-1)`
`k_(1)=(2.303)/(k)"log"((100)/(100-99))`
`0.1=(2.303)/(t)"log"(100)/(1)`
`0.1=(2.303)/(t)log 10^(2)`
`:. t=(2.303xx2)/(0.1)=46.06 min`.