Chapter 17 - Radioactivity and Nuclear Chemistry - Exercises - Problems - Page 638: 72

45.6 days

Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{11.4\,d}=0.06079\,d^{-1}$ Original amount $A_{0}=0.240\,mol$ Amount remaining $A=1.50\times10^{-2}\,mol$ Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required. $\implies \ln(\frac{0.240\,mol}{1.50\times10^{-2}\,mol})=2.77259=0.06079\,d^{-1}(t)$ $\implies t=\frac{2.77259}{0.06079\,d^{-1}}=45.6\,d$