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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 _ _ _ × 10 - 3 min - 1 . (Nearest integer) [Use : ln 10 = 2 . 303 ; log 10 3 = 0 . 477 property of logarithm : log x y = y log x

Correct answer

0

Step-by-step solution

As the unit of rate constant is min - 1 so it must be a first order reaction k × t = 2 . 303 logA 0 / A t in 1 min 10 % is inactivated so taking A 0 = 100      A t = 90 in 1   min So k × 1 = 2 . 303 × log 100 90 = 2 . 303 × ( log 10 - 2 log 3 ) = 2 . 303 × ( 1 - 2 × 0 . 477 ) = 0 . 10593 = 105 . 93 × 10 - 3 Answer is 106

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