$\begin{array}{1 1} 3.12 \% \\ 6.24\;\% \\ 1.56\% \\ 9.36\;\% \end{array}$

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Answer: 3.12%

From the integrated rate law for a first order reaction, $ [A] = [A]_0\; e^{-kt}$.

The half life $t_{1/2} = \large\frac{ ln\;2}{k}$$ = \large\frac{0.693}{k}$. Given $t_{1/2} = 180\;s \rightarrow k = \large\frac{0.693}{180\;s}$ $ = 0.00385 s^{-1}$

The fraction remaining is the concentration divided by the initial concentration $ = \large\frac{[A]}{[A_0]}$$ = e^{kt} = e^{0.00385 s^{-1} \times 900\; s}$$ = 0.0312 = 3.12\%$

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