The inactivation rate of a viral preparation in 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___________ × 10^-3min^-1. (Nearest integer)

[Use: $l n 10$ = 2.303; log₁₀ 3 = 0.477; property of logarithm: log x^y = y log x]

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alok kumar singh | Contributor-Level 10

2 months ago

Kt = 2.303 log $\frac{[A_{0}]}{[A_{t}]}$

Here, $[A_{0}] = 100, [A_{t}] = 90 a n d t = 1 m i n$

$∴ K \times 1 = 2.303 l o g_{10} \frac{100}{90}$

$∴ K = 2.303 [1 - 2 \times 0.477] = 0.105938 = 105.938 \times 1 0^{- 3} m i n^{- 1}$

Rounded = 106

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