Answer:
10.88 %
Explanation:
Given that:
Half life = 15.0 hours
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{15.0}\ hour^{-1}[/tex]
The rate constant, k = 0.04621 hour⁻¹
Time = 48.0 hours
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
So,
[tex]\frac {[A_t]}{[A_0]}=e^{-0.04621\times 48}[/tex]
[tex]\frac {[A_t]}{[A_0]}=0.10881[/tex]
In percentage, the sample of sodium-24 remains = 10.88 %
