Question

The half-life of iodine-131 is about 8 days. How much of a 50mg sample will be left in 25 days? Write your answer rounded to the nearest tenth.

Answer 1

Answer:

5.73 mg of the sample will be left in 25 days.

Explanation:

Given that:

Half life = 8 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,  

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{ln\ 2}{8}\ days^{-1}$

The rate constant, k = 0.08664 days⁻¹

Time = 25 days

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration  = 50 mg

So,  

$[A_t]=50\times e^{-0.08664\times 25}\ mg$

$[A_t]=5.73\ mg$

5.73 mg of the sample will be left in 25 days.