Answer:
[tex][A_t]=54.5\ g[/tex]
Explanation:
Given that:
Half life = 14.0 days
[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}{14.0}\ days^{-1}[/tex]
The rate constant, k = 0.04951 days⁻¹
Initial concentration [A₀] = 60.0 g
Time = 46.7 hrs
Considering, 1 hr = 0.041667 days
So, time = 1.9458 days
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
So,
[tex][A_t]=60.0\times e^{-0.04951\times 1.9458}\ g[/tex]
[tex][A_t]=54.5\ g[/tex]
