A first order reaction has a rate constant 1.15 x 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 3 g? 

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{R}}_0}{\mathrm{R}}$
Here, $\begin{array}{r}\mathrm{K}=1.15\times {10}^{-3}{\mathrm{s}}^{-1}\\ {\mathrm{R}}_0=5\mathrm{gm},\mathrm{r}=3\mathrm{gm}\end{array}$
$\begin{array}{l}\therefore 1.15\times {10}^{-3}=\frac{2.303}{\mathrm{t}}\log \frac{5}{3}\\ =\frac{2.303}{\mathrm{t}}[\log 5-\log 3]\\ 1.15\times {10}^{-3}=\frac{2.303}{\mathrm{t}}[0.699-0.477]\\ =\frac{2.303}{\mathrm{t}}\times 0.222\end{array}$
$\begin{array}{l}\therefore \mathrm{t}=\frac{2.303\times 0.222}{1.15\times {10}^{-3}}\sec \\ =\frac{2.303\times 0.222}{1.15}\times 1000\\ =445\sec \\ =7\min .41\sec \end{array}$