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A first order reaction has a rate constant of $2.303\times10^{-3}$ s$^{-1}$. The time required for 40 g of this reactant to reduce to 10 g will be [Given that $\log_{10}2=0.3010$]
Detailed Solution
$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{2.303\times10^{-3}}=301$ s. 40 g to 10 g is two half-lives (75% completion): $t=2\times301=602$ s.