A reaction has a half life of 1min . The time required for 99 . 9 %...

\dfrac{{t}_{99 . 9 %}}{{t}_{50 %}} = \dfrac{\dfrac{1}{K}ln\dfrac{100}{0 . 1}}{\dfrac{1}{K}\ln 2}

= \dfrac{\ln 1000}{\ln 2} \times {t}_{50 %}

$= \dfrac{3\ln 10}{\ln 2} \times 1$

$= \dfrac{3 \times 2 . 3}{0 . 69} = 10$