A first order reaction has a rate constant of $2.303\times {10}^{-3}{\mathrm{s}}^{-1}$. The time required for 40 g of this reactant to reduce to 10 g will be $\text{ [Given that }\left.{\log }_{10}2=0.3010\right]$

For a first order reaction

Half life period, $\begin{array}{r}{\mathrm{t}}_{\frac{1}{2}}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{2.303\times {10}^{-3}}{\mathrm{s}}^{-1}\\ =300.91\mathrm{s}\end{array}$

$\text{ Now, }40\mathrm{g}\stackrel{{\mathrm{t}}_{1/2}}{\to } 20\mathrm{g}\stackrel{{\mathrm{t}}_{1/2}}{\to } 10\mathrm{g}$

So, 40 g substance requires 2 half life periods
to reduce up to 10 g
$\therefore$ Time taken in reduction = 2 × 300.91 s 

$\begin{array}{l} =601.82\\ \simeq 602\mathrm{s}\end{array}$