$A$ $20 \%$ first-order reaction is completed in $32 \ \text{min}$. How much time (in $\text{min}$) will it take for $60 \%$ of the reaction to be completed (in $.00$)?

An organic compound undergoes first order decomposition. The time taken for decomposition to $\left(\frac{1}{8}\right)^{\text{th}}$ and $\left(\frac{1}{10}\right)^{\text{th}}$ 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: $\log 2 = 0.3$)

For a first-order reaction,the time taken for the initial concentration to decrease to $1/4$ of its value is $20 \ min$. The time taken for the initial concentration to decrease to $1/16$ of its value will be $......... \ min$.

The half-life of a first-order reaction is $10 \ minutes$. If the initial concentration is $0.08 \ mol/L$ and at some instant the concentration is $0.01 \ mol/L$,then the time elapsed is ...... minutes.

For the first order reaction $A \rightarrow B$,the half-life is $30 \ min$. The time taken for $75 \%$ completion of the reaction is $..... \ min$. (Nearest integer)
Given: $\log 2 = 0.3010, \log 3 = 0.4771, \log 5 = 0.6989$

For a first-order reaction,which of the following statements is correct?