Show that in case of a first order reaction, the time required for `99.9%` of the reaction to take place is about `10` times that the required for half the reaction.

Explanation

For reaction of first order, `t_(1//2)=(2.303)/(k)log""(a)/(a-(a)/(2))=(2.303)/(k)log""2=(2.303)/(k)(0.3010)`
`t_(99.9%=(2.303)/(k)log=(a)/(a-0.999a)=(2.303)/(k)log10^(-3)=(2.303)/(k)xx3:.(t_(99.9%))/(t_(1//2))=(3)/(0.3010)=10`