A first-order reaction has a half-life of 32 min. If the initial concentration of the reactant is 2.7 x 1015 molecules/L, what is the concentration after 2.24 hours?

The given time is $2.24 \text{ h}$. Since $1 \text{ h} = 60 \text{ min}$, we convert:

$$t = 2.24 \text{ h} \times \frac{60 \text{ min}}{1 \text{ h}} = 134.4 \text{ min}$$

For a first-order reaction, the half-life ($t_{1/2}$) is related to the rate constant ($k$) by the formula:

$$t_{1/2} = \frac{\ln(2)}{k}$$

Given $t_{1/2} = 32 \text{ min}$:

$$k = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{32 \text{ min}} \approx 0.02166 \text{ min}^{-1}$$

The integrated rate law for a first-order reaction is:

$$\ln([A]_t) - \ln([A]_0) = -kt$$

Rearranging to solve for $[A]_t$:

$$\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$$ $$\frac{[A]_t}{[A]_0} = e^{-kt}$$ $$[A]_t = [A]_0 \cdot e^{-kt}$$

Given $[A]_0 = 2.7 \times 10^{15} \text{ molecules/L}$, $k = 0.02166 \text{ min}^{-1}$, and $t = 134.4 \text{ min}$:

$$[A]_t = (2.7 \times 10^{15} \text{ molecules/L}) \cdot e^{-(0.02166 \text{ min}^{-1} \times 134.4 \text{ min})}$$ $$[A]_t = (2.7 \times 10^{15} \text{ molecules/L}) \cdot e^{-2.907}$$ $$[A]_t = (2.7 \times 10^{15} \text{ molecules/L}) \cdot 0.05462$$ $$[A]_t \approx 1.4747 \times 10^{14} \text{ molecules/L}$$

Rounding to two significant figures:

$$[A]_t \approx \boxed{\text{1.5} \times 10^{14} \text{ molecules/L}}$$