Answer :
The governing equation for a continuous (exponential) decay is given by,
[tex]y=y_0\cdot e^{kt}[/tex]Here, the amount of substance at the beginning (t=0) is,
[tex]y_0[/tex](a)
Given that the half-life is 4 minutes,
[tex]\begin{gathered} y=\frac{y_0}{2} \\ y_0\cdot e^{k(4)}=\frac{y_0}{2} \\ e^{4k}=\frac{1_{}}{2} \\ \ln (e^{4k})=\ln (\frac{1}{2}) \\ 4k=\ln (2)^{-1} \\ 4k=-\ln (2) \\ k=\frac{-1}{4}\ln (2) \end{gathered}[/tex]Substitute the value of 'k' in the expression,
[tex]y=y0\cdot e^{(\frac{-1}{4}\ln (2))t}[/tex]It is mentioned that the amount is 118.4 gm at the beginning of the experiment,
[tex]y_0=118.4[/tex]Then the equation becomes,
[tex]y=118.4e^{(\frac{-1}{4}\ln (2))t}[/tex]This is the required formula relating 'y' to 't'.
(b)
The amount corresponding to 15 minutes is calculated as,
[tex]\begin{gathered} y=118.4e^{(\frac{-1}{4}\ln 2)(15)} \\ y\approx118.4(0.074) \\ y\approx8.8 \end{gathered}[/tex]Thus, the amount of substance in 15 minutes is 8.8 gm approximately.