Question

3. A 5 gm/100 ml solution of drug X is stored in a closed test tube

Answer 1

Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

Where;

ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

ln0.5 = ln5 -0.05t

t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

t= 46 days