Q.

$$\begin{gathered} {\mathrm{t}}_{1/2}=45 \text{minutes} \\ \text{For 1st order reaction }{\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}} \\ \therefore \mathrm{k}=\frac{0.693}{45}{\min }^{-1}=0.0154{\min }^{-1} \\ \text{ Again for 1st order reaction we have } \\ \mathrm{k}=\frac{2.303}{\mathrm{t}}\log \left[\frac{100}{100-99.9}\right] \\ \text{ (considering original concentration to be 100) } \\ \begin{array}{rl} & \mathrm{t}=\frac{2.303}{\mathrm{k}}\log \left[\frac{100}{0.1}\right]\\ & \mathrm{t}=\frac{2.303}{0.0154}\log 1000\\ & \mathrm{t}=\frac{2.303}{0.0154}\times 3=448.636 \text{mintes}\end{array} \\ \mathrm{t}=7\frac{3}{10}-=7\frac{1}{2}\mathrm{hours} \end{gathered}$$