The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, $10 \%$ of the virus is inactivated. The rate constant for viral inactivation is ________ $\times 10^{-3} \mathrm{~min}^{-1} \cdot$ (Nearest integer)
$\left[\right.$ Use $: \ln 10=2.303 ; \log _{10} 3=0.477 ;$
property of logarithm: $\left.\log x^{y}=y \log x\right]$
As the unit of rate constant is $\mathrm{min}^{-1}$ so it must be a first order reaction
$\mathrm{K} \times \mathrm{t}=2.303 \log \mathrm{A}_{0} / \mathrm{A}_{\mathrm{t}}$
in $1 \mathrm{~min} 10 \%$ is in activated so tabing
$\mathrm{A}_{0}=100 \quad \mathrm{~A}_{\mathrm{t}}=90$ in $1 \mathrm{~min}$
So $\mathrm{K} \times 1=2.303 \times \log \frac{100}{90}$
$=2.303 \times(\log 10-2 \log 3)$
$=2.303 \times(1-2 \times 0.477)$
$=0.10593$
$=105.93 \times 10^{-3}$
$\approx 106$