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Q.
Half life of a first order reaction is $4 \,s$ and the initial concentration of the reactant is $0.12 \,M$. The concentration of the reactant left after $16 \,s$ is
Chemical Kinetics
Solution:
$t_{1 / 2}=4 s T=16 s$
$n=\frac{T}{t_{1 / 2}}=\frac{16}{4}=4$
$\left(\therefore T=n \times t_{1 / 2}\right)$
$[A]=\left[A_{0}\right]\left(\frac{1}{2}\right)^{n}$
$=0.12 \times\left(\frac{1}{2}\right)^{4}=\frac{0.12}{16}=0.0075\,M$
Where $\left[A_{0}\right]=$ initial concentration and $[A]$
$=$ concentration left after time $t$.