Answer:
Approximately [tex]4.5\; \text{hours}[/tex].
Explanation:
Calculate the ratio between the mass of this sample after [tex]22.5\; \text{hours}[/tex] and the initial mass:
[tex]\displaystyle \frac{5.70\; \rm g}{182.4\; \rm g} \approx 0.03125[/tex].
Let [tex]n[/tex] denote the number of half-lives in that [tex]22.5\; \text{hours}[/tex] (where [tex]n\![/tex] might not necessarily be an integer.) The mass of the sample is supposed to become [tex](1/2)[/tex] the previous quantity after each half-life. Therefore, if the initial mass of the sample is [tex]1\; \rm g[/tex] (for example,) the mass of the sample after [tex]\! n[/tex] half-lives would be [tex]{(1/2)}^{n}\; \rm g[/tex]. Regardless of the initial mass, the ratio between the mass of the sample after [tex]n\!\![/tex] half-lives and the initial mass should be [tex]{(1/2)}^{n}[/tex].
For this question:
[tex]{(1/2)}^{n}} = 0.03125[/tex].
Take the natural logarithm of both sides of this equation to solve for [tex]n[/tex]:
[tex]\ln \left[{(1/2)}^{n}}\right] = \ln (0.03125)[/tex].
[tex]n\, [\ln(1/2)] = \ln (0.03125)[/tex].
[tex]\displaystyle n = \frac{\ln(0.03125)}{\ln(1/2)} \approx 5[/tex].
In other words, there are [tex]5[/tex] half-lives of this sample in [tex]22.5\; \text{hours}[/tex]. If the length of each half-life is constant, that length should be [tex](1/5) \times 22.5\; \text{hours} = 4.5\; \text{hours}[/tex].
