The half-life period of a first order reaction is 600 s. What percent of A remains after 30 minutes ?

Answer

`12.5%`

Explanation

`600" s"=10" min, "300" min "=3" half lives, i.e.,"n=3.` Hence, `%" of A left "=(1)/(2^(n))[A]_(0)=(1)/(2^(3))[A]_(0)=12.5%.`