Chapter 9, Problem 82P

Interpretation Introduction

Interpretation: The half-life of phosphorus-32 is 14.3 days. The initial amount of sample of radioactive phosphorus-32 is 1.2 mg. The time given is 28.6 days. The amount of initial sample remaining after 28.6 days needs to be determined.

Concept Introduction:The integrated rate law for first order reaction is represented as follows:

$k=\frac{2.303}{t}\log \frac{\left[A_o\right]}{\left[A_t\right]}$

Here, k is rate constant, $\left[A_o\right]$ is initial amount and $\left[A_t\right]$ is final amount of the sample.