$\begin{array}{1 1} 1110 \;s \\ 1100\;s \\ 1081\;s \\ 1006\;s \end{array}$

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Answer: 1110 s

Before we calculate the time we need to calculate $k$ the rate constant first.

Given 32.5% of the reactant remains

$ln\;A = -kt + ln\;A_0 \rightarrow ln 0.325 = -k \times 540\;s + ln \; 1 \rightarrow k = 0.00208s^{-1}$

Going back to the integrated form of the rate law for first order reactions: $ln\;A = -kt + ln\;A_0$

$\Rightarrow ln\; 0.10 = - (0.0020813 s^{-1}) (t) + ln 1 \rightarrow t \approx 1110 \;s$

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