Answer: The mass of sample that remained is 0.127 mg
Explanation:
The integrated rate law equation for first-order kinetics:
[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(1)
Given values:
a = initial concentration of reactant = 0.500 mg
a - x = concentration of reactant left after time 't' = ?mg
t = time period = 28 s
k = rate constant = [tex]0.049s^{-1}[/tex]
Putting values in equation 1:
[tex]0.049s^{-1}=\frac{2.303}{28s}\log (\frac{0.500}{(a-x)})\\\\\log (\frac{0.500}{(a-x)})=\frac{0.049\times 28}{2.303}\\\\\frac{0.500}{a-x}=10^{0.5957}\\\\frac{0.500}{a-x}=3.94\\\\a-x=\frac{0.500}{3.942}=0.127mg[/tex]
Hence, the mass of sample that remained is 0.127 mg