A first order reaction has a rate constant of 2.303 × 10^ -3 ~s^ -1...

For first order reaction, $\mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$Given : $\mathrm{k}=2.303 \times 10^{-3} \mathrm{ s}^{-1}, \mathrm{a}=40 \mathrm{ g}, \mathrm{a}-\mathrm{x}=10 \mathrm{ g}On substituting the$given values in Eq. (i), we get $$\begin{aligned}t & =\frac{2.303}{2.303 \times 10^{-3}} \log \frac{40}{10}\\ & =10^3 \log 2^2=2 \times10^3 \times \log2 \\ & =2 \times10^3 \times0.3010=602 \mathrm{ s} \end{aligned}$$$Alternative method For first order reaction,$\mathrm{t}_{1/2}=\frac{0.693}{\mathrm{k}}\mathrm{t}_{1/2}\left(\mathrm{t}_{50 \%}\right)=\frac{0.693}{2.303 \times 10^{-3}}=$301$\mathrm{ s}$Also,$\quad \mathrm{t}_{75 \%}=2 \mathrm{t}_{50 \%} \therefore \quad \mathrm{t}_{75 \%}=$2$\times$301=602$\mathrm{ s}