If half-life of a substance is 36 minutes. Find amount left after 2...

Given, $\mathrm{t}_{1/2}=36 \mathrm{ min}, \mathrm{t}=2 \mathrm{hrs}=120 \mathrm{ min},[\mathrm{Ao}]=\log$ For first order reaction, $\mathrm{R}=\frac{0.693}{\mathrm{t}_{1/2}}$$\begin{aligned} \therefore \quad k & =\frac{0.693}{36}=0.01925 \\ t & =\frac{2303}{k} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]} \end{aligned}$where, $\left[\mathrm{A}_0\right]=$ initial amount\begin{aligned} & {[\mathrm{A}]=\text{ final amount } } \\ & 120=\frac{2303}{0.01925} \log \frac{10}{[\mathrm{ A}]} \\ & \log 10-\log [\mathrm{A}]=1 \\ & 1-\log [\mathrm{A}]=1 \\ & \log [\mathrm{A}]=0 \\ & {[\mathrm{ A}] }=1 \mathrm{gm} \end{aligned}$$