The time taken for 90% of a first order reaction to be completed is approximately

Options

A. 1.1 times that of half-life
B. 2.2 times that of half-life
C. 3.3 times that of half-life
D. 4.4 times that of half-life

Answer

C

Explanation

`t_(90%)=(2.303)/(k)log(100)/(100-90)" "(I)`
`t_(50%)=(2.303)/(k)log(100)/(100-50)" "(II)`
Dividing `(t_(90%))/(t_(50%))=(log 10)/(log 2)`
`:. t_(90%)=3.3 t_(50%)`