Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, the amount of reactant remaining after $n$ half-lives is given by $N_t=\frac{N_0}{2^n}$

Given initial amount $N_0=10$ g and final amount $N_t=2.5$ g

Substituting the values, $2.5=\frac{10}{2^n}$

$2^n=\frac{10}{2.5}=4$

$2^n=2^2\Rightarrow n=2$

Total time required $t=n\times t_{1/2}$

Given $t_{1/2}=5$ hour

$t=2\times 5=10$ hour

Answer: $10$ hour