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  • 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 __________  <img alt="\times 10^{-3} \text{min}^{

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} \text{min}^{-1}. (Nearest integer) \mathrm{[Use : \ln 10=2.303 ; log _{10} 3=0.477 ; property \; of\; logarithm : \log \mathrm{x}^{\mathrm{y}}=\mathrm{y} \log \mathrm{x}]}
 

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best_answer

Given,

\text{rate}\; \alpha\; [\text{virus}]

The inactivation follows first order kinetics.

Given, C_{t}=\frac{90}{100} C_{0} \Rightarrow \frac{C_{0}}{C_{t}}=\frac{100}{90}

From first order kinetics,

\begin{aligned} &\therefore \ln \left(\frac{C_{0}}{C_{t}}\right)=k t . \\ &\Rightarrow \ln \left(\frac{100}{90}\right)=k(1) \\ &\Rightarrow k=\ln \left(\frac{100}{90}\right) \end{aligned}

\begin{aligned} &=2.303 \log \left(\frac{10}{9}\right) \\ &=2.303(\log 10-\log 9) \\ &=2.303(1-2 \log 3) \\ &=2.303(1-0.954) \\ &=0.106 \text {min} ^{-1} \end{aligned}

=106 \times 10^{-3} \mathrm{~min}^{-1} .

Hence, the correct answer is 106.

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