Half life of a first order reaction in 10 min. What `%` of reaction will be completed in 100 min ? `
A. 0.25
B. 0.5
C. 0.999
D. 0.75

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1 Answers

Correct Answer - C
`(0.693)/(t_(1//2)) = (2.303)/(t) "log" (a)/(a - x)`
`(0.693)/(10) = (2.303)/(100) "log" (100)/(100 - x) implies x = 99.9%`

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