Answer:
[tex]t=147.24\ min[/tex]
Explanation:
Given that:
Half life = 30 min
[tex]t_{1/2}=\frac{\ln2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac{\ln2}{t_{1/2}}[/tex]
[tex]k=\frac{\ln2}{30}\ min^{-1}[/tex]
The rate constant, k = 0.0231 min⁻¹
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
Given that:
The rate constant, k = 0.0231 min⁻¹
Initial concentration [tex][A_0][/tex] = 7.50 mg
Final concentration [tex][A_t][/tex] = 0.25 mg
Time = ?
Applying in the above equation, we get that:-
[tex]0.25=7.50e^{-0.0231\times t}[/tex]
[tex]750e^{-0.0231t}=25[/tex]
[tex]750e^{-0.0231t}=25[/tex]
[tex]x=\frac{\ln \left(30\right)}{0.0231}[/tex]
[tex]t=147.24\ min[/tex]
