Question

Sodium-24 is used to treat leukemia and has a half-life of 15 hours. A patient was injected with a salt solution containing sodium-24. What percentage of the 24na remained after 48 hours?

Answer 1

10.88 %

Step-by-step explanation:

Given that:

Half life = 15.0 hours

`t_(1/2)=\frac {ln\ 2}{k}`

Where, k is rate constant

So,

`k=\frac {ln\ 2}{t_(1/2)}`

`k=\frac {ln\ 2}{15.0}\ hour^(-1)`

The rate constant, k = 0.04621 hour⁻¹

Time = 48.0 hours

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

So,

`\frac {[A_t]}{[A_0]}=e^(-0.04621* 48)`

`\frac {[A_t]}{[A_0]}=0.10881`

In percentage, the sample of sodium-24 remains = 10.88 %