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(B) Let $N_0 = 1000 	ext{ molecules}/m^3$ be the initial concentration.In one minute,the number of molecules leaving is $\frac{1}{10} N_0$. Thus,the

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$70$ minutes

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(B) Let $N_0 = 1000 	ext{ molecules}/m^3$ be the initial concentration.In one minute,the number of molecules leaving is $\frac{1}{10} N_0$. Thus,the remaining concentration after $1 	ext{ minute}$ is $N_1 = N_0 - \frac{1}{10} N_0 = \frac{9}{10} N_0 = 0.9 N_0$.This follows the discrete decay model $N_t = N_0(0.9)^t$,where $t$ is in minutes.We want to find $t$ such that $N_t = 1 	ext{ molecule}/m^3$.$1 = 1000(0.9)^t$$(0.9)^t = 0.001$Taking the natural logarithm on both sides:$t \ln(0.9) = \ln(0.001)$$t \approx \frac{-6.9077}{-0.10536} \approx 65.56 	ext{ minutes}$.Rounding to the nearest provided option,$t \approx 70 	ext{ minutes}$.

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