An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1/8$ and $1/10$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log_{10} 2 = 0.3$)

The reaction,$N_2O_5 \longrightarrow 2NO_2 + \frac{1}{2} O_2$ is of first order for $N_2O_5$ with rate constant $6.2 \times 10^{-4} \ s^{-1}$. What is the value of rate of reaction when $[N_2O_5] = 1.25 \ mol \ L^{-1}$?

In a first order reaction,the concentration of the reactant decreases from $20 \ mmol$ to $10 \ mmol$ in $1.151 \ min$. What is the rate constant (in $min^{-1}$)?

$N_{2}O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$
In the above first order reaction,the initial concentration of $N_{2}O_{5}$ is $2.40 \times 10^{-2} \ mol \ L^{-1}$ at $318 \ K$. The concentration of $N_{2}O_{5}$ after $1 \ hour$ was $1.60 \times 10^{-2} \ mol \ L^{-1}$. The rate constant of the reaction at $318 \ K$ is $..... \times 10^{-3} \ min^{-1}$. (Nearest integer)
[Given: $\log 3 = 0.477, \log 5 = 0.699$]

If $60\%$ of a first order reaction was completed in $60 \ min$,$50\%$ of the same reaction would be completed in approximately $.......... \ min.$
$(\log \, 4 = 0.60, \, \log \, 5 = 0.69)$

$A$ flask is filled with equal moles of $A$ and $B$. The half-lives of $A$ and $B$ are $100 \, s$ and $50 \, s$ respectively and are independent of the initial concentration. The time required for the concentration of $A$ to be four times that of $B$ is $.... \, s.$
(Given : $\ln 2 = 0.693$ )