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

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

Step-by-step 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=(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* e^(-0.08664* 25)\ mg`

`[A_t]=5.73\ mg`

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