Question

what is the answer of question(f)

Answer 1

In first order reaction the equation is :::> A=Ao(e^(-kt) where k is the rate constant

A IS THE AMOUNT OF SUBSTANCE LEFT AFTER THE REACTION

so for case 1 it is A= (1/10)*Ao thus solving we get t=ln(10)/k  [note::ln(1/10)= - ln(10)]

for case 2 its is A=(1/100)Ao thus getting t=ln(100)/k=2*ln(10)/k  [ln(100)=2*ln(10)]

HENCE PROVED